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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 $0$

$\frac{13}{237}-1\cdot 0arctan\left(8\cdot 0\right)+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
25

Simplifying

$\frac{13}{237}+0arctan\left(0\right)+\lim_{c\to0}\left(\frac{1}{16}\ln\left|1+64c^2\right|\right)$
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 $0$

$0.0549+0.0625\ln\left|1+64\left(0^2\right)\right|$
28

Simplifying

$0.0549+0.0625\ln\left|1\right|$
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 $0$

$\frac{13}{237}$

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