VERKKOKIRJA

Remark. This definition of the integral is sometimes referred to as the Darboux integral.

For non-negative functions $f$ this definition of the integral coincides with the idea of making the difference between the the areas of the circumscribed and the inscribed rectangles arbitrarily small by using ever finer partitions.

Proof.

Remark. This definition of the integral turns out to be equivalent to that of the Darboux integral i.e. a function is Riemann-integrable if and only if it is Darboux-integrable and the values of the two integrals are always equal.

Remark. Any interval $[a,b]$ can be partitioned into equally spaced subintervals by setting $\Delta x = (b-a)/n$ and $x_{k} = a + k\Delta x$.

Piecewise-defined functions

Example

Consider the function $f\colon\left[-1,1\right]$ defined as $f(x) = \begin{cases} -1 &\text{ for }-1\le x$ We can now integrate $f$ as follows: $\int_{-1}^{1}f(x)\,dx = \int_{-1}^{0}f(x)\,dx + \int_{0}^{1}f(x)\,dx$ $=\int_{-1}^{0}(-1)\,dx + \int_{0}^{1}1\,dx = -1\cdot(-1-0) + 1\cdot(1-0) = 2.$

Important properties

Fundamental theorem of calculus

Proof.

Proof.

If $F'(x)=f(x)$ on some open interval then $F$ is the antiderivative (or the primitive function) of $f$. The fundamental theorem of calculus guarantees that for every continuous function $f$ there exists an antiderivative $F(x) = \int_{a}^{x}f(t)\,dt.$ The antiderivative is not necessarily expressible as a combination of elementary functions even if $f$ were an elementary function, e.g. $f(x) = e^{-x^{2}}$. Such primitives are called nonelementary antiderivatives.

Proof.

Proof.

Proof.

Constant Functions

Given the constant function $f(x) = c,\,c\in\mathbb{R}$. The integral $\int\limits_a^b f(x)\,\mathrm{d} x = \int\limits_a^b c \, \mathrm{d}x$ has to be determined now.

Solution by finding a antiderivative

From the previous chapter it is known that $g(x) = c\cdot x$ gives $g'(x) = c$. This means that $c \cdot x$ is an antiderivative for $c$. So the following applies $\int\limits_a^b c \, \mathrm{d}x = [c \cdot x]_{x=a}^{x=b} = c\cdot b - c \cdot a = c \cdot (b-a).$

Remark: Of course, a function $h(x) = c \cdot x + d$ would also be an antiderivative of $f$, since the constant $d$ is omitted in the derivation. For sake of simplicity $c \cdot x$ can be used, since $d$ can be chosen as $d=0$ for definite integrals.

Solution by geometry

The area under the constant function forms a rectangle with height $c$ and length $b-a$. Thus the area is $c \cdot (b-a)$ and this corresponds to the solution of the integral. Illustrate this remark by a sketch.

Linear functions

Given is the linear function $f(x) = mx$. We are looking for the integral $\int\limits_a^b f(x)\, \mathrm dx=\int\limits_a^b mx\, \mathrm dx$.

Solve by finding a antiderivative

The antiderivative of a linear function is in any case a quadratic function, since $\frac{\mathrm d x^2}{\mathrm dx} = 2x$. The derivative of a quadratic function results in a linear function. Here, it is important to consider the leading factor as in $\frac{\mathrm d (m \cdot \frac{1}{2} \cdot x^2)}{\mathrm dx} = mx.$ Thus the result is $\int\limits_a^b mx \mathrm{d}x = \left[\frac{m}{2}x^2 \right]_{x=a}^{x=b}= \frac{m}{2}b^2 - \frac{m}{2}a^2.$

Solving by geometry

The integral $\int\limits_a^b mx\, \mathrm dx$ can be seen geometrically, as subtracting the triangle with the edges $(0|0)$, $(a|0)$ and $(a| ma)$ from the triangle with the edges $(0|0)$, $(b|0)$ and $(b| mb)$. Since the area of a triangle ist given by $\frac{1}{2} \cdot \mbox{baseline} \cdot \mbox{height}$, the area of the first triangle $\frac{1}{2}\cdot b \cdot mb = \frac{1}{2}mb^2$ and that of the second triangle is analogous $\frac{1}{2}ma^2$. For the integral the result is $\frac{m}{2}b^2 - \frac{m}{2}a^2$. This is consistent with the integral calculated using the antiderivative. Illustrate this remark by a sketch.

Power functions

In constant and linear functions we have already seen that the exponent of a function decreases by one when it is derived. So it has to get bigger when integrating. The following applies: $\frac{\mathrm d x^n}{\mathrm dx} = n \cdot x^{n-1}.$ It follows that the antiderivative for $x^n$ must have the exponent $n+1$, $\frac{\mathrm d x^{n+1}}{\mathrm dx} = (n+1) \cdot x^n.$ By multiplying the last equation with $\frac{1}{n+1}$ we get $\frac{\mathrm d}{\mathrm dx}\frac{1}{n+1} x^{n+1} = \frac{n+1}{n+1} \cdot x^n = x^n.$ Finally the antiderivative is $\int x^n\, \mathrm dx = \frac{1}{n+1} x ^{n+1}+c,\,c\in\mathbb{R}$.

Examples

• $\int x^2\, \mathrm dx = \frac{1}{3} x^3 +c,\,c\in\mathbb{R}$
• $\int x^3\, \mathrm dx = \frac{1}{4} x^4 +c,\,c\in\mathbb{R}$
• $\int x^{20}\, \mathrm dx = \frac{1}{21} x^{21} +c,\,c\in\mathbb{R}$

The formula $\int x^n\, \mathrm dx = [\frac{1}{n+1} x^{n+1}]$ is also valid, if the exponent of the function is a real number and not equal $-1$.

Examples

• $\int x^{2,7}\, \mathrm dx = \frac{1}{3,7} x^{3,7}+c,\,c\in\mathbb{R}$
• $\int \sqrt{x}\, \mathrm dx = \int x^\frac{1}{2} = \frac{2}{3} x^{\frac{3}{2}}+c,\,c\in\mathbb{R}$
• But: For $x\gt0$ applies$\int x^{-1}\,\mathrm d x=\ln(x)+c,\,c\in\mathbb{R}.$

Natural Exponential function

The natural exponential function $f(x) = e^x$ is one of the easiest function to differentiate and integrate. Since the derivation of $e^x$ results in $e^x$, it follows $\int e^x\, \mathrm dx = e^x +c, \, c\in \mathbb{R}.$

Example 1

Determine the value of the integral $\int_0^1 e^z \,\mathrm{d} z$.

$\int\limits_0^1e^z\,\mathrm{d}z= e^z\big|_{z=0}^{z=1}=e^1-e^0=e-1.$

Example 2

Determine the value of the integral $\int_0^b e^{\alpha t} \,\mathrm{d} t$. Using the same considerations as above we get $\int\limits_0^b e^{\alpha t}\,\mathrm{d}t= \frac{1}{\alpha}e^{\alpha t}\big|_{t=0}^{t=b} =\frac{1}{\alpha}\left(e^{\alpha b}-e^0\right)=\frac{1}{\alpha}\left(e^{\alpha b}-1\right).$ Important is here, that we have to use the factor $\frac1{\alpha}$.

Natural Logarithm

The derivative of the natural logarithmic function is $\ln'(x) =\frac{1}{x}$ for $x\gt0$. It even applies $\ln'(x) =\frac{1}{x}$ to $x$. These results together result in for the antiderivative of $\frac{1}{x}$

$\int \frac{1}{x}\,\mathrm{d}x = \ln\left(|x|\right) +c , c\in\mathbb{R}.$

An antiderivative can be specified for the natural logarithm: $\int \ln(x)\,\mathrm{d}x = x\ln(x) - x + c ,\, c\in\mathbb{R}.$

Trigonometric function

The antiderivatives of $\sin(x)$ and $\cos(x)$ also result logically if you derive "backwards". We have $\int \sin(x)\, \mathrm dx = -\cos(x)+c,\,c\in\mathbb{R},$ since $(-\cos(x))' =-(-\sin(x))=\sin(x).$ Furthermore we know $\int \cos(x)\, \mathrm dx = \sin(x)+c,\,c\in\mathbb{R},$ since $(\sin(x))' = \cos(x)$ applies.

Example 1

Which area is covered by the sine on the interval $[0,\pi]$ and the $x$-axis? To determination the area we simply have to evaluate the integral $\int_0^{\pi} \sin(\tau) \, \mathrm{\tau}.$ That means $\int_0^{\pi} \sin(\tau) \, \mathrm{d}\tau = \left[-\cos(\tau)\right]_{\tau=0}^{\tau=\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2.$ Again make a sketch for this example.

Example 2

How can the integral $\int \cos(\omega t +\phi)\,\mathrm{d}t$ be expressed analytically?

To determine the integral we use the antiderivative of the cosine: $\sin'(x) = \cos(x)$. However, the inner derivativ has to be considered in the given function and thus we get $\int \ \cos(\omega t +\phi)\,\mathrm{d}t =\frac{1}{\omega}\sin(\omega t+\phi)+c,\,c\in\mathbb{R}.$

Example 1

Evaluate the integrals $\displaystyle \int_{-1}^{1}e^{-x}\,dx$ and $\displaystyle\int_{0}^{1}\sin(\pi x)\,dx$.

Solution. The antiderivative of $e^{-x}$ is $-e^{-x}$ so we have that $\int_{-1}^{1}e^{-x}\,dx = -e^{-1}+e^{1} = 2\sinh1.$ The antiderivative of $\sin(\pi x)$ is $-\frac{1}{\pi}\cos(\pi x)$ and thus $\int_{0}^{1}\sin(\pi x)\,dx = -\frac{1}{\pi}(\cos\pi-\cos0) = \frac{2}{\pi}.$

Example 2

Evaluate the integral $\displaystyle\int_{0}^{1}\frac{x}{\sqrt{25-9x^{2}}}\,dx$.

Solution. The antiderivative might look something like $F(x)=a(25-9x^{2})^{1/2}$, where we can find the factor $a$ through differentiation: $D\big(a(25-9^{2})^{1/2}\big) = a\cdot\frac{1}{2}\cdot(-18x)(25-9x^{2})^{-1/2} = \frac{-9ax}{\sqrt{25-9x^{2}}}$ hence if $a=-1/9$ we get the correct antiderivative. Thus $\int_{0}^{1}\frac{x}{\sqrt{25-9x^{2}}}\,dx = -\frac{1}{9}\cdot(25-9x^{2})^{1/2}\Big|_{x=0}^{x=1} = -\frac{1}{9}(\sqrt{16}-\sqrt{25}) = \frac{1}{9}.$ This integral can also be solved using integration by substitution; more on this method later.

Suppose that $f$ and $g$ are piecewise continuous functions. The area of a region bounded by the graphs $y=f(x)$, $y=g(x)$ and the vertical lines $x=a$ and $x=b$ is given by the integral $A=\int_{a}^{b}|f(x)-g(x)|\,dx.$

Especially if $f$ is a non-negative function on the interval $[a,b]$ and $g(x)=0$ for all $x$ then the integral $A=\int_{a}^{b}f(x)\,dx$ is the area of the region bounded by the graph $y=f(x)$, the $x$-axis and the vertical lines $x=a$ and $x=b$.

The arc length $\ell$ of a planar curve $y=f(x)$ between points $x=a$ and $x=b$ is given by the integral $\ell = \int_{a}^{b}\sqrt{1+f'(x)^{2}}\,dx.$

Heuristic reasoning: On a small interval $\left[x,x+\Delta x\right]$ the arc length of the curve between $y=f(x)$ and $y=f(x+\Delta x)$ is approximately $\Delta s \approx \sqrt{\Delta x^{2} + \Delta y^{2}} = \Delta x\sqrt{1+\left(\frac{\Delta y}{\Delta x}\right)^{2}} \approx \Delta x\sqrt{1+f'(x)^{2}}.$

The area of a surface generated by rotating the graph $y=f(x)$ around the $x$-axis on the interval $\left[a,b\right]$ is given by $A = 2\pi\int_{a}^{b}|f(x)|\sqrt{1+f'(x)^{2}}\,dx.$ Heuristic reasoning: An area element of the surface is approximately $\Delta A \approx \text{perimeter}\cdot\text{length} = 2\pi|f(x)|\cdot\Delta s.$

Solid of revolution

Suppose that the cross-sectional area of a solid is given by the function $A(x)$ when $x\in\left[a,b\right]$. Then the volume of the solid is given by the integral $V = \int_{a}^{b}A(x)\,dx.$ If the graph $y=f(x)$ is rotated around the $x$-axis between the lines $x=a$ and $x=b$ the volume of the generated figure (the solid of revolution) is $V = \pi\int_{a}^{b}f(x)^{2}\,dx.$ This follows from the fact that the cross-sectional area of the figure at $x$ is a circle with radius $f(x)$ i.e. $A(x)=\pi f(x)^{2}$.

More generally: Let $0\le g(x)\le f(x)$ and suppose that the region bounded by $y=f(x)$ and $y=g(x)$ and the lines $x=a$ and $x=b$ is rotated around the $x$-axis. The volume of this solid of revolution is $V = \pi\int_{a}^{b}\big(f(x)^{2}-g(x)^{2}\big)\,dx.$

One limitation of the improper integration is that the limit must be taken with respect to one endpoint at a time.

Example

$\int_{0}^{\infty}\frac{dx}{\sqrt{x}(1+x)} = \int_{0}^{1}\frac{dx}{\sqrt{x}(1+x)} + \int_{1}^{\infty}\frac{dx}{\sqrt{x}(1+x)}$ Provided that both of the integrals on the right-hand side converge. If either of the two is divergent then so is the integral.

Example

Find the value of $\displaystyle\int_{0}^{\infty}e^{-x}\,dx$.

Solution. Notice that $\int_{0}^{R}e^{-x}\,dx = \left(-e^{-x}\right)\Bigg|_{x=0}^{R}=1-e^{-R}\to 1$ as $R\to\infty$. Thus the improper integral converges and $\int_{0}^{\infty}e^{-x}\,dx = 1.$

However, this doesn't apply in general. For example, let $f(x)=x$. Note that even though $\int_{-R}^{R}f(x)\,dx = \int_{-R}^{R}x\,dx = \frac{R^2}{2} - \frac{(-R)^2}{2} = 0$ for all $R\in\mathbb{R}$ the improper integral $\int_{-\infty}^{\infty}x\,dx = \lim_{R\to\infty}\int_{-R}^{0}x\,dx + \lim_{R\to\infty}\int_{0}^{R}x\,dx = \lim_{R\to\infty}-\frac{(-R)^2}{2} + \lim_{R\to\infty}\frac{R^2}{2} = \infty - \infty$ does not converge.

Example

Find the value of the improper integral $\displaystyle\int_{0}^{1}\frac{dx}{\sqrt{x}}$.

Solution. We get $\int_{\epsilon}^{1}\frac{dx}{\sqrt{x}} = \left(2\sqrt{x}\right)\Bigg|_{x=\epsilon}^{x=1} = 2-2\sqrt{\epsilon} \to 2,$ as $\epsilon\to0+$. Thus the integral converges and its value is $2$.

Integration techniques

Logarithmic integration

Given a quotient of differentiable functions, we know to apply the quotient rule. However, this is not so easy with integration. Here only for a few special cases we will state rules in this chapter.

Logarithmic integration As we already know the derivative of $\ln(x)$, i.e. the natural logarithm to the base $e$, equal to $\frac{1}{x}$. According to the chain rule the derivative of differentiable function with positive function values is $f\,:\,\frac{\mathrm d}{\mathrm dx} \ln (f(x)) = \frac{f'(x)}{f(x)}$. This means that for a quotient of functions where the numerator is the derivative of the denominator yields the rule: \begin{equation} \int \frac{f'(x)}{f(x)}\, \mathrm{d} x= \ln \left(|f(x)|\right) +c,\,c\in\mathbb{R}.\end{equation} Using the absolute value of the function is important, since the logarithm is defined on $\mathbb{R}^+$.

Examples

• $\int \frac{1}{x}\, \mathrm dx = \int \frac{x'}{x} \mathrm\, dx = \ln(|x|) +c,\,c\in\mathbb{R}$.
  
  
• $\int \frac{3x^2 + 17}{x^3 +17x - 15}\, \mathrm dx = \ln(|x^3 + 17x - 15|)+c,\,c\in\mathbb{R}$.
  
  
• $\int \frac{\cos(x)}{\sin(x)}\,\mathrm dx = \ln(|\sin(x)|)+c,\,c\in\mathbb{R}$.

Integration of rational functions - partial fraction decomposition

The logarithmic integration works well in special cases of broken rational functions where the counter is a multiple of the derivation of the denominator. However, other cases can sometimes be traced back to this. This method is called partial fractional decomposition, which represents rational functions as the sum of proper rational functions.

Example 1

The function $\frac{1}{1-x^2}$ cannot be integrated at first glance. However, the denominator $1-x^2$ can be written as $(1-x)(1+x)$ and the function can finally reads as $\dfrac{1}{1-x^2} = \dfrac{\frac{1}{2}}{1+x} + \dfrac{\frac{1}{2}}{1-x}$ by partial fraction decomposition. This expression can be integrated, as demonstrated now: \begin{eqnarray} \int \dfrac{1}{1-x^2} \,\mathrm dx &= & \int \dfrac{\frac{1}{2}}{1+x} + \dfrac{\frac{1}{2}}{1-x}\, \mathrm dx \\ & =& \frac{1}{2} \int \dfrac{1}{1+x}\, \mathrm dx - \frac{1}{2} \int \dfrac{-1}{1-x}\, \mathrm dx\\ & = &\frac{1}{2} \ln|1+x| +c_1 - \frac{1}{2} \ln|1-x| +c_2\\ &= &\frac{1}{2} \ln \left|\dfrac{1+x}{1-x}\right|+c,\,c\in\mathbb{R}. \end{eqnarray} This procedure is now described in more detail for some special cases.

Case 1: $Q(x)=(x-\lambda_1)(x-\lambda_2)$ with $\lambda_1\ne\lambda_2$. In this case, $R$ has the representation $R(x) = \frac{ax+b}{(x-\lambda_1)(x-\lambda_2)}$ and can be transformed to $\frac{ax+b}{(x-\lambda_1)(x-\lambda_2)} = \frac{A}{(x-\lambda_1)}+\frac{B}{(x-\lambda_2)}.$ By multiplying with $(x-\lambda_1)(x-\lambda_2)$ it yields ot $ax+b = A(x-\lambda_1) + B(x-\lambda_2) = \underbrace{(A+B)}_{\stackrel{!}{=}a}x + \underbrace{(-A\lambda_1-B\lambda_2)}_{\stackrel{!}{=}b}.$

$A$ and $B$ are now obtained by the method of equating the coefficients.

Example 2

Determe the partial fraction decomposition of $\frac{2x+3}{(x-4)(x+5)}$.

Start with the equation $\frac{2x+3}{(x-4)(x+5)} = \frac{A}{(x-4)}+\frac{B}{(x+5)}$ to get the parameters $A$ and $B$. Multiplication by ${(x-4)(x+5)}$ leads to $2x+3 = A(x+5)+B(x-4) = (A+B)x +5A -4B.$ Now we get the system of linear equations

\begin{eqnarray}A+B & = & 2 \\ 5A - 4 B &=& 3\end{eqnarray} with the solution $A = \frac{11}{9}$ and $B= \frac{7}{9}$. The representation with proper rational functions is $\frac{2x+3}{(x-4)(x+5)}=\frac{11}{9}\frac{1}{(x-4)}+\frac{7}{9}\frac{1}{(x+5)}$ The integral of the type $\int \frac{ax+b}{(x-\lambda_1)(x-\lambda_2)}\,\mathrm{d} x.$ is no longer mystic.

With the help of partial fraction decomposition, this integral can now be calculated in the following manner \begin{eqnarray}\int \frac{ax+b}{(x-\lambda_1)(x-\lambda_2)}\mathrm{d} x &=& \int\frac{A}{(x-\lambda_1)}+\frac{B}{(x-\lambda_2)}\mathrm{d} x \\ &=&A\int\frac{1}{(x-\lambda_1)}\mathrm{d} x +B\int\frac{1}{(x-\lambda_2)}\mathrm{d} x \\ & = & A\ln(|x-\lambda_1|) + B\ln(|x-\lambda_2|).\end{eqnarray}

Example 3

Determine the antiderivative for $\frac{2x+3}{(x-4)(x+5)}$, i.e. $\int\frac{2x+3}{(x-4)(x+5)}\,\mathrm{d} x.$

From the above example we already know: $\int\frac{2x+3}{(x-4)(x+5)}\,\mathrm{d} x = \int\frac{11}{9}\frac{1}{(x-4)}+\frac{7}{9}\frac{1}{(x+5)}\, \mathrm{d} x.$

Using the idea explained above immediately follow: $\int\frac{11}{9}\frac{1}{(x-4)}+\frac{7}{9}\frac{1}{(x+5)} \,\mathrm{d} x = \frac{11}{9}\int\frac{1}{(x-4)} \mathrm{d} x + \frac{7}{9}\int\frac{1}{(x+5)} \,\mathrm{d} x= \frac{11}{9}\ln(|(x-4)|)+\frac{7}{9}\ln(|(x+5)|).$ So is the result $\int\frac{2x+3}{(x-4)(x+5)}\,\mathrm{d} x=\frac{11}{9}\ln(|(x-4)|)+\frac{7}{9}\ln(|(x+5)|).$

Case 2: $Q(x)=(x-\lambda)^2$.

In this case $R$ has the representation $R(x) = \frac{ax+b}{(x-\lambda)^2}$ and the ansatz $\frac{ax+b}{(x-\lambda)^2} = \frac{A}{(x-\lambda)}+\frac{B}{(x-\lambda)^2}$ is used.

By multiplying the equation with $(x-\lambda)^2$ we get $ax+b = A(x-\lambda) + B.$ Again equating the coefficients leads us to a system of linear equations in $A=a$ and $B=b+A\lambda=b+a\lambda.$

So we have $\int \frac{ax+b}{(x-\lambda)^2}\,\mathrm{d}x = \int \frac{a}{(x-\lambda)}+\frac{b+a\lambda}{(x-\lambda)^2} \mathrm{d}x =a\ln(|x-\lambda|)-\frac{b+a\lambda}{(x-\lambda)}+c,\,c\in\mathbb{R}.$

3. case $Q(x)=x^2+mx+n$ without real zeros.

In this case $R$ has the representation $R(x) = \frac{ax+b}{x^2+mx+n}$ and the representation can not be simplified.

Only the special case $R(x) = \frac{2x+m}{x^2+mx+n}$ is now considered.

In this case we have $\int \frac{2x+m}{x^2+mx+n}\,\mathrm{d}x = \ln(|x^2+mx+n|)+c, \quad c\in\mathbb{R}.$

Another special case is $R(x) = \frac{1}{x^2+1}$ with $\int \frac{1}{x^2+1} \, \mathrm{d} x = \arctan(x) +c,\quad c\in \mathbb{R}.$

Integration by Parts

The derivative of a product of two continuously differentiable functions $f$ and $g$ is $(f(x)\cdot g(x))' = f'(x)\cdot g(x)+f(x)\cdot g'(x),\quad x\in(a,b).$

This leads us to the following theorem:

Proof

Example

Solve the integral $\displaystyle\int_{0}^{\pi}x\sin x\,dx$.

Solution. Set $f'(x)=\sin x$ and $g(x) = x$. Then $f(x)=-\cos x$ and $g'(x) = 1$ and the integration by parts gives $\int_{0}^{\pi}x\sin x\,dx = -\pi\cos\pi - 0 - \int_{0}^{\pi}(-\cos x)\,dx$ $=\pi+\left(\sin x\right)\Bigg|_{x=0}^{\pi} = \pi.$

Notice that had we chosen $f$ and $g$ the other way around this would have led to an even more complicated integral.

Integration by Substitution

Proof.

Example 1

Find the value of the integral $\displaystyle\int_{0}^{\pi^2}\sin\sqrt{x}\,dx$.

Solution. Making the substitution $x=t^{2}$ when $t\ge0$ we have $dx=2t\,dt$. Solving the limits from the inverse formula i.e. $t=\sqrt{x}$ we find that $t(0)=0$ and $t(\pi^{2})=\pi$. Hence $\int_{0}^{\pi^{2}}\sin\sqrt{x}\,dx = \int_{0}^{\pi^{2}}2t\sin t\,dt = 2\int_{0}^{\pi}t\sin t\,dt = 2\pi.$

Here the latter integral was solved applying integration by parts in the previous example.

Example 2

Find the antiderivative of $\displaystyle\frac{1}{\sqrt{x}(1+x)}$.

Solution. Substituting $x=t^{2}$, $t>0$ or $t=\sqrt{x}$ gives $\int\frac{dx}{\sqrt{x}(1+x)} = \int\frac{2t}{t(1+t^{2})}\,dt = 2\arctan t + C = 2\arctan\sqrt{x} + C.$