1
Solved example of improper integrals
$\int_0^{\left(\frac{1}{8}\right)}\left(\arctan\left(8x\right)\right)dx$
2
Apply integration by parts: $u=arctan(8x)$, $v'=1$
$\left(xarctan\left(8x\right)-\int_{0}^{\frac{1}{8}}\frac{8x}{1+64x^2}dx\right)$
3
Taking the constant out of the integral
$\left(xarctan\left(8x\right)-8\int_{0}^{\frac{1}{8}}\frac{x}{1+64x^2}dx\right)$
4
Solve the integral $\int\frac{x}{1+64x^2}dx$ applying u-substitution. Let $u$ and $du$ be
$\begin{matrix}u=1+64x^2 \\ du=128xdx\end{matrix}$
5
Isolate $dx$ in the previous equation
$\frac{du}{128x}=dx$
6
Substituting $u$ and $dx$ in the integral and simplify
$\left(xarctan\left(8x\right)-8\int_{0}^{\frac{1}{8}}\frac{1}{128u}du\right)$
7
Replace the integral's limit by a finite value
$\lim_{c\to0}\:\left(xarctan\left(8x\right)-8\int_{c}^{\frac{1}{8}}\frac{1}{128u}du\right)$
8
Take the constant out of the integral
$\lim_{c\to0}\:\left(xarctan\left(8x\right)-\frac{1}{16}\int_{c}^{\frac{1}{8}}\frac{1}{u}du\right)$
9
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
$\lim_{c\to0}\left(\left[\left(xarctan\left(8x\right)-\frac{1}{16}\ln\left|u\right|\right)\right]_{c}^{\frac{1}{8}}\right)$
10
Substitute $u$ back for it's value, $1+64x^2$
$\lim_{c\to0}\left(\left[\left(xarctan\left(8x\right)-\frac{1}{16}\ln\left|1+64x^2\right|\right)\right]_{c}^{\frac{1}{8}}\right)$
11
Evaluate the definite integral
$\lim_{c\to0}\left(\frac{1}{8}arctan\left(8\cdot \frac{1}{8}\right)-\frac{1}{16}\ln\left|1+64\left(\frac{1}{8}^2\right)\right|-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
12
Multiply $8$ times $\frac{1}{8}$
$\lim_{c\to0}\left(\frac{1}{8}arctan\left(1\right)-\frac{1}{16}\ln\left|1+64\left(\frac{1}{8}^2\right)\right|-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
13
Calculate the power $0.125^2$
$\lim_{c\to0}\left(\frac{1}{8}arctan\left(1\right)-\frac{1}{16}\ln\left|1+64\cdot \frac{1}{64}\right|-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
14
Calculating the arctangent of $1$
$\lim_{c\to0}\left(\frac{1}{8}\cdot \frac{2}{\sqrt{2}}-\frac{1}{16}\ln\left|1+64\cdot \frac{1}{64}\right|-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
15
Multiply $64$ times $\frac{1}{64}$
$\lim_{c\to0}\left(\frac{27}{275}-\frac{1}{16}\ln\left|1+1\right|-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
16
Add the values $1$ and $1$
$\lim_{c\to0}\left(\frac{27}{275}-\frac{1}{16}\ln\left|2\right|-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
17
Calculating the absolute value of $2$
$\lim_{c\to0}\left(\frac{27}{275}-\frac{1}{16}\ln\left(2\right)-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
18
Calculating the natural logarithm of $2$
$\lim_{c\to0}\left(\frac{27}{275}-\frac{1}{16}\cdot \frac{4}{\sqrt{3}}-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
19
Multiply $-\frac{1}{16}$ times $\frac{4}{\sqrt{3}}$
$\lim_{c\to0}\left(\frac{27}{275}-\frac{12}{277}-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
20
Subtract the values $\frac{27}{275}$ and $-\frac{12}{277}$
$\lim_{c\to0}\left(\frac{13}{237}-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)\right)$
21
Solve the product $-\left(carctan\left(8c\right)-\frac{1}{16}\ln\left|1+64c^2\right|\right)$
$\lim_{c\to0}\left(\frac{13}{237}-carctan\left(8c\right)+\frac{1}{16}\ln\left|1+64c^2\right|\right)$
22
The limit of a sum of two functions is equal to the sum of the limits of each function: $\displaystyle\lim_{x\to c}(f(x)\pm g(x))=\lim_{x\to c}(f(x))\pm\lim_{x\to c}(g(x))$
$\lim_{c\to0}\left(\frac{13}{237}\right)+\lim_{c\to0}\left(-carctan\left(8c\right)\right)+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
23
The limit of a constant is just the constant
$\frac{13}{237}+\lim_{c\to0}\left(-carctan\left(8c\right)\right)+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
24
Evaluating the limit when $c$ tends to $0$
$\frac{13}{237}-1\cdot 0arctan\left(8\cdot 0\right)+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
$\frac{13}{237}+0arctan\left(0\right)+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
26
Any expression multiplied by $0$ is equal to $0$
$\frac{13}{237}+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
27
Evaluating the limit when $c$ tends to $0$
$0.0549+0.0625\ln\left|1+64\left(0^2\right)\right|$
$0.0549+0.0625\ln\left|1\right|$
29
Calculating the absolute value of $1$
$0.0549+0.0625\ln\left(1\right)$
30
Calculating the natural logarithm of $1$
$0.0549+0.0625\cdot 0$
31
Any expression multiplied by $0$ is equal to $0$
$\frac{13}{237}$