# Integration by trigonometric substitution Calculator

## Get detailed solutions to your math problems with our Integration by trigonometric substitution step by step calculator. Sharpen your math skills and learn step by step with our math solver. Check out more online calculators here.

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### Difficult Problems

1

Solved example of Integration by substitution

$\int\frac{2x+1}{\left(x-1\right)^3\left(x^2+4\right)^2}dx$
2

Rewrite the fraction $\frac{2x+1}{\left(x-1\right)^3\left(x^2+4\right)^2}$ in $5$ simpler fractions using partial fraction decomposition

$\frac{2x+1}{\left(x-1\right)^3\left(x^2+4\right)^2}=\frac{A}{\left(x-1\right)^3}+\frac{Bx+C}{\left(x^2+4\right)^2}+\frac{D}{x-1}+\frac{F}{\left(x-1\right)^{2}}+\frac{Gx+H}{x^2+4}$
3

Find the values of the unknown coefficients. The first step is to multiply both sides of the equation by $\left(x-1\right)^3\left(x^2+4\right)^2$

$2x+1=\left(x-1\right)^3\left(x^2+4\right)^2\left(\frac{A}{\left(x-1\right)^3}+\frac{Bx+C}{\left(x^2+4\right)^2}+\frac{D}{x-1}+\frac{F}{\left(x-1\right)^{2}}+\frac{Gx+H}{x^2+4}\right)$
4

Multiplying polynomials

$2x+1=\frac{A\left(x-1\right)^3\left(x^2+4\right)^2}{\left(x-1\right)^3}+\frac{\left(x-1\right)^3\left(x^2+4\right)^2\left(Bx+C\right)}{\left(x^2+4\right)^2}+\frac{D\left(x-1\right)^3\left(x^2+4\right)^2}{x-1}+\frac{F\left(x-1\right)^3\left(x^2+4\right)^2}{\left(x-1\right)^{2}}+\frac{\left(x-1\right)^3\left(x^2+4\right)^2\left(Gx+H\right)}{x^2+4}$
5

Simplifying

$2x+1=A\left(x^2+4\right)^2+\left(x-1\right)^3\left(Bx+C\right)+D\left(x-1\right)^{2}\left(x^2+4\right)^2+F\left(x^2+4\right)^2\left(x-1\right)+\left(x-1\right)^3\left(x^2+4\right)\left(Gx+H\right)$
6

Expand the polynomial

$2x+1=A\left(x^2+4\right)^2+\left(x-1\right)^3\left(Bx+C\right)+D\left(x-1\right)^{2}\left(x^2+4\right)^2+\left(x^2+4\right)^2\left(Fx-F\right)+\left(x-1\right)^3\left(x^2+4\right)\left(Gx+H\right)$
7

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}-1=25A-8\left(-B+C\right)+100D-50F-40\left(-G+H\right)&\:\:\:\:\:\:\:(x=-1) \\ 3=25A&\:\:\:\:\:\:\:(x=1) \\ -7=400A-125\left(-4B+C\right)+10000D-2000F-2500\left(-4G+H\right)&\:\:\:\:\:\:\:(x=-4) \\ 9=400A+27\left(4B+C\right)+3600D+1200F+540\left(4G+H\right)&\:\:\:\:\:\:\:(x=4) \\ -9=841A-216\left(-5B+C\right)+30276D-5046F-6264\left(-5G+H\right)&\:\:\:\:\:\:\:(x=-5) \\ 11=841A+64\left(5B+C\right)+13456D+3364F+1856\left(5G+H\right)&\:\:\:\:\:\:\:(x=5) \\ 13=1600A+125\left(6B+C\right)+40000D+8000F+5000\left(6G+H\right)&\:\:\:\:\:\:\:(x=6)\end{matrix}$
8

Proceed to solve the system of linear equations

$\begin{matrix}25A & - & 1B & + & 1C & + & 100D & - & 50F & - & 1G & + & 1H & =-1 \\ 25A & + & 0B & + & 0C & + & 0D & + & 0F & + & 0G & + & 0H & =3 \\ 400A & - & 4B & + & 1C & + & 10000D & - & 2000F & - & 4G & + & 1H & =-7 \\ 400A & + & 4B & + & 1C & + & 3600D & + & 1200F & + & 4G & + & 1H & =9 \\ 841A & - & 5B & + & 1C & + & 30276D & - & 5046F & - & 5G & + & 1H & =-9 \\ 841A & + & 5B & + & 1C & + & 13456D & + & 3364F & + & 5G & + & 1H & =11 \\ 1600A & + & 6B & + & 1C & + & 40000D & + & 8000F & + & 6G & + & 1H & =13\end{matrix}$
9

Rewrite as a coefficient matrix

$\left(\begin{matrix}25 & -1 & 1 & 100 & -50 & -1 & 1 & -1 \\ 25 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \\ 400 & -4 & 1 & 10000 & -2000 & -4 & 1 & -7 \\ 400 & 4 & 1 & 3600 & 1200 & 4 & 1 & 9 \\ 841 & -5 & 1 & 30276 & -5046 & -5 & 1 & -9 \\ 841 & 5 & 1 & 13456 & 3364 & 5 & 1 & 11 \\ 1600 & 6 & 1 & 40000 & 8000 & 6 & 1 & 13\end{matrix}\right)$
10

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{3}{25} \\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & -243.5608 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & -961.7216 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -\frac{1}{66} \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & -\frac{13}{142} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 270.0557 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 981.1546\end{matrix}\right)$
11

The decomposed integral equivalent is

$\int\left(\frac{\frac{3}{25}}{\left(x-1\right)^3}+\frac{-243.5608x-961.7216}{\left(x^2+4\right)^2}+\frac{-\frac{1}{66}}{x-1}+\frac{-\frac{13}{142}}{\left(x-1\right)^{2}}+\frac{270.0557x+981.1546}{x^2+4}\right)dx$
12

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int\frac{\frac{3}{25}}{\left(x-1\right)^3}dx+\int\frac{-243.5608x-961.7216}{\left(x^2+4\right)^2}dx+\int\frac{-\frac{1}{66}}{x-1}dx+\int\frac{-\frac{13}{142}}{\left(x-1\right)^{2}}dx+\int\frac{270.0557x+981.1546}{x^2+4}dx$
13

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|, where b=-1 and n=-\frac{1}{66} \int\frac{\frac{3}{25}}{\left(x-1\right)^3}dx+\int\frac{-243.5608x-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\int\frac{-\frac{13}{142}}{\left(x-1\right)^{2}}dx+\int\frac{270.0557x+981.1546}{x^2+4}dx 14 Apply the formula: \int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(x+a\right)^{\left(c-1\right)}\left(c-1\right)}$, where $a=-1$, $c=3$ and $n=\frac{3}{25}$

$\frac{-\frac{3}{25}}{2\left(x-1\right)^{2}}+\int\frac{-243.5608x-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\int\frac{-\frac{13}{142}}{\left(x-1\right)^{2}}dx+\int\frac{270.0557x+981.1546}{x^2+4}dx$
15

Apply the formula: $\frac{a}{bx}$$=\frac{\frac{a}{b}}{x}, where a=-\frac{3}{25}, b=2 and x=\left(x-1\right)^{2} \frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\int\frac{-243.5608x-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\int\frac{-\frac{13}{142}}{\left(x-1\right)^{2}}dx+\int\frac{270.0557x+981.1546}{x^2+4}dx 16 Apply the formula: \int\frac{n}{\left(x+a\right)^c}dx$$=\frac{-n}{\left(x+a\right)^{\left(c-1\right)}\left(c-1\right)}$, where $a=-1$, $c=2$ and $n=-\frac{13}{142}$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\int\frac{-243.5608x-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+\int\frac{270.0557x+981.1546}{x^2+4}dx$
17

Split the fraction $\frac{-243.5608x+-961.7216494845361}{\left(x^2+4\right)^2}$ in two terms with same denominator

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\int\left(\frac{-243.5608x}{\left(x^2+4\right)^2}+\frac{-961.7216}{\left(x^2+4\right)^2}\right)dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+\int\frac{270.0557x+981.1546}{x^2+4}dx$
18

Split the fraction $\frac{270.0557x+981.1546391752578}{x^2+4}$ in two terms with same denominator

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\int\left(\frac{-243.5608x}{\left(x^2+4\right)^2}+\frac{-961.7216}{\left(x^2+4\right)^2}\right)dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+\int\left(\frac{270.0557x}{x^2+4}+\frac{981.1546}{x^2+4}\right)dx$
19

The integral of a sum of two or more functions is equal to the sum of their integrals

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\int\frac{-243.5608x}{\left(x^2+4\right)^2}dx+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+\int\frac{270.0557x}{x^2+4}dx+\int\frac{981.1546}{x^2+4}dx$
20

Taking the constant out of the integral

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-243.5608\int\frac{x}{\left(x^2+4\right)^2}dx+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+\int\frac{270.0557x}{x^2+4}dx+\int\frac{981.1546}{x^2+4}dx$
21

Taking the constant out of the integral

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-243.5608\int\frac{x}{\left(x^2+4\right)^2}dx+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+\int\frac{981.1546}{x^2+4}dx$
22

Take the constant out of the integral

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-243.5608\int\frac{x}{\left(x^2+4\right)^2}dx+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+981.1546\int\frac{1}{4+x^2}dx$
23

Factor the integral's denominator by $4$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-243.5608\int\frac{x}{\left(x^2+4\right)^2}dx+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+245.2887\int\frac{1}{1+\frac{1}{4}x^2}dx$
24

Solve the integral $\int\frac{x}{\left(x^2+4\right)^2}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x^2+4 \\ du=2xdx\end{matrix}$
25

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
26

Substituting $u$ and $dx$ in the integral and simplify

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-243.5608\int\frac{1}{2u^2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+245.2887\int\frac{1}{1+\frac{1}{4}x^2}dx$
27

Take the constant out of the integral

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-121.7804\int\frac{1}{u^2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+245.2887\int\frac{1}{1+\frac{1}{4}x^2}dx$
28

Solve the integral applying the substitution $v^2=\frac{1}{4}x^2$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-121.7804\int\frac{1}{u^2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+2245.2887\int\frac{1}{1+v^2}dv$
29

Multiply $2$ times $245.28865979381445$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-121.7804\int\frac{1}{u^2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773\int\frac{1}{1+v^2}dv$
30

Solve the integral applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-121.7804\int\frac{1}{u^2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(v\right)$
31

Substitute $v$ back for it's value, $\frac{1}{2}x$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-121.7804\int\frac{1}{u^2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
32

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}-121.7804\int u^{-2}du+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
33

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+121.7804u^{-1}+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
34

Substitute $u$ back for it's value, $x^2+4$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+121.7804\left(x^2+4\right)^{-1}+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
35

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int\frac{-961.7216}{\left(x^2+4\right)^2}dx-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
36

Solve the integral $\int\frac{-961.7216}{\left(x^2+4\right)^2}dx$ by trigonometric substitution using the substitution

$\begin{matrix}x=2\tan\left(\theta\right) \\ dx=2\sec\left(\theta\right)^2d\theta\end{matrix}$
37

Substituting in the original integral, we get

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int\frac{-1923.4433\sec\left(\theta\right)^2}{\left(4\tan\left(\theta\right)^2+4\right)^2}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
38

Factor by the greatest common divisor $4$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int\frac{-1923.4433\sec\left(\theta\right)^2}{\left(4\left(\tan\left(\theta\right)^2+1\right)\right)^2}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
39

The power of a product is equal to the product of it's factors raised to the same power

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int\frac{-1923.4433\sec\left(\theta\right)^2}{16\left(\tan\left(\theta\right)^2+1\right)^2}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
40

Apply the formula: $\frac{ab}{cf}$$=\frac{a}{c}\cdot\frac{b}{f}, where a=-1923.4433, b=\sec\left(\theta\right)^2, c=16 and f=\left(\tan\left(\theta\right)^2+1\right)^2 \frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int-120.2152\left(\frac{\sec\left(\theta\right)^2}{\left(\tan\left(\theta\right)^2+1\right)^2}\right)d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right) 41 Applying the trigonometric identity: \tan(x)^2+1=\sec(x)^2 \frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int-120.2152\left(\frac{\sec\left(\theta\right)^2}{\sec\left(\theta\right)^{4}}\right)d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right) 42 Simplifying the fraction by \sec\left(\theta\right) \frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int-120.2152\sec\left(\theta\right)^{\left(2-4\right)}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right) 43 Subtract the values 2 and -4 \frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int-120.2152\sec\left(\theta\right)^{-2}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right) 44 Apply the formula: \sec\left(x\right)^n$$=\cos\left(x\right)^{-n}$, where $x=\theta$ and $n=-2$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}+\int-120.2152\cos\left(\theta\right)^{2}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
45

Take the constant out of the integral

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\int\cos\left(\theta\right)^{2}d\theta-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
46

Apply the formula: $\int\cos\left(x\right)^2dx$$=\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)$, where $x=\theta$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}\theta+\frac{1}{4}\sin\left(2\theta\right)\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
47

Expressing the result of the integral in terms of the original variable

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{1}{4}\sin\left(2\theta\right)\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
48

Using the sine double-angle identity

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(\theta\right)\cos\left(\theta\right)\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{x}{x^2+4}dx+490.5773arctan\left(\frac{1}{2}x\right)$
49

Solve the integral $\int\frac{x}{x^2+4}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x^2+4 \\ du=2xdx\end{matrix}$
50

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
51

Substituting $u$ and $dx$ in the integral and simplify

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(\theta\right)\cos\left(\theta\right)\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{1}{2u}du+490.5773arctan\left(\frac{1}{2}x\right)$
52

Expressing the result of the integral in terms of the original variable

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{x}{x^2+4}\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+270.0557\int\frac{1}{2u}du+490.5773arctan\left(\frac{1}{2}x\right)$
53

Take the constant out of the integral

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{x}{x^2+4}\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+135.0278\int\frac{1}{u}du+490.5773arctan\left(\frac{1}{2}x\right)$
54

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{x}{x^2+4}\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+135.0278\ln\left|u\right|+490.5773arctan\left(\frac{1}{2}x\right)$
55

Substitute $u$ back for it's value, $x^2+4$

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{x}{x^2+4}\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+135.0278\ln\left|x^2+4\right|+490.5773arctan\left(\frac{1}{2}x\right)$
56

As the integral that we are solving is an indefinite integral, when we finish we must add the constant of integration

$\frac{-\frac{3}{50}}{\left(x-1\right)^{2}}+\frac{121.7804}{x^2+4}-120.2152\left(\frac{1}{2}arctan\left(\frac{x}{2}\right)+\frac{x}{x^2+4}\right)-\frac{1}{66}\ln\left|x-1\right|+\frac{\frac{13}{142}}{x-1}+135.0278\ln\left|x^2+4\right|+490.5773arctan\left(\frac{1}{2}x\right)+C_0$

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