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Integration by substitution Calculator

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1

Solved example of Combining like terms

$\int\:\frac{-x^2+8x^2-9x+2}{\left(x^2+1\right)\left(x-3\right)^2}dx$
2

Adding $-1x^2$ and $8x^2$

$\int\frac{-9x+2+7x^2}{\left(x-3\right)^2\left(x^2+1\right)}dx$
3

Rewrite the fraction $\frac{-9x+2+7x^2}{\left(x-3\right)^2\left(x^2+1\right)}$ in $3$ simpler fractions using partial fraction decomposition

$\frac{-9x+2+7x^2}{\left(x-3\right)^2\left(x^2+1\right)}=\frac{Ax+B}{x^2+1}+\frac{C}{\left(x-3\right)^2}+\frac{D}{x-3}$
4

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

$-9x+2+7x^2=\left(x-3\right)^2\left(x^2+1\right)\left(\frac{Ax+B}{x^2+1}+\frac{C}{\left(x-3\right)^2}+\frac{D}{x-3}\right)$
5

Multiplying polynomials

$-9x+2+7x^2=\frac{\left(x-3\right)^2\left(x^2+1\right)\left(Ax+B\right)}{x^2+1}+\frac{C\left(x-3\right)^2\left(x^2+1\right)}{\left(x-3\right)^2}+\frac{D\left(x-3\right)^2\left(x^2+1\right)}{x-3}$
6

Simplifying

$-9x+2+7x^2=\left(x-3\right)^2\left(Ax+B\right)+C\left(x^2+1\right)+D\left(x^2+1\right)\left(x-3\right)$
7

Expand the polynomial

$-9x+2+7x^2=A\left(x-3\right)^2x+B\left(x-3\right)^2+Cx^2+C+Dx^{3}-3Dx^2+Dx-3D$
8

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

$\begin{matrix}18=-16A+16B-3D-3D+2C-2D&\:\:\:\:\:\:\:(x=-1) \\ 0=4A+4B+2C+2D-6D&\:\:\:\:\:\:\:(x=1) \\ 92=-108A+36B+9C+C-27D-3D-30D&\:\:\:\:\:\:\:(x=-3) \\ 38=10C&\:\:\:\:\:\:\:(x=3)\end{matrix}$
9

Proceed to solve the system of linear equations

$\begin{matrix} -16A & + & 16B & + & 2C & - & 8D & =18 \\ 4A & + & 4B & + & 2C & - & 4D & =0 \\ -108A & + & 36B & + & 10C & - & 60D & =92 \\ 0A & + & 0B & + & 10C & + & 0D & =38\end{matrix}$
10

Rewrite as a coefficient matrix

$\left(\begin{matrix}-16 & 16 & 2 & -8 & 18 \\ 4 & 4 & 2 & -4 & 0 \\ -108 & 36 & 10 & -60 & 92 \\ 0 & 0 & 10 & 0 & 38\end{matrix}\right)$
11

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & 0 & 0 & -\frac{51}{50} \\ 0 & 1 & 0 & 0 & \frac{7}{50} \\ 0 & 0 & 1 & 0 & \frac{19}{5} \\ 0 & 0 & 0 & 1 & \frac{51}{50}\end{matrix}\right)$
12

The decomposed integral equivalent is

$\int\left(\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}+\frac{\frac{19}{5}}{\left(x-3\right)^2}+\frac{\frac{51}{50}}{x-3}\right)dx$
13

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

$\int\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}dx+\int\frac{\frac{19}{5}}{\left(x-3\right)^2}dx+\int\frac{\frac{51}{50}}{x-3}dx$
14

Apply the formula: $\int\frac{n}{x+b}dx$$=n\ln\left|x+b\right|$, where $b=-3$ and $n=\frac{51}{50}$

$\int\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}dx+\int\frac{\frac{19}{5}}{\left(x-3\right)^2}dx+\frac{51}{50}\ln\left|x-3\right|$
15

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 $a=-3$, $c=2$ and $n=\frac{19}{5}$

$\int\frac{-\frac{51}{50}x+\frac{7}{50}}{x^2+1}dx+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
16

Split the fraction $\frac{-\frac{51}{50}x+0.14}{x^2+1}$ in two terms with same denominator

$\int\left(\frac{-\frac{51}{50}x}{x^2+1}+\frac{\frac{7}{50}}{x^2+1}\right)dx+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
17

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

$\int\frac{-\frac{51}{50}x}{x^2+1}dx+\int\frac{\frac{7}{50}}{x^2+1}dx+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
18

Taking the constant out of the integral

$-\frac{51}{50}\int\frac{x}{x^2+1}dx+\int\frac{\frac{7}{50}}{x^2+1}dx+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
19

Take the constant out of the integral

$-\frac{51}{50}\int\frac{x}{x^2+1}dx+\frac{7}{50}\int\frac{1}{1+x^2}dx+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
20

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

$-\frac{51}{50}\int\frac{x}{x^2+1}dx+\frac{7}{50}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
21

Solve the integral $\int\frac{x}{x^2+1}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x^2+1 \\ du=2xdx\end{matrix}$
22

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
23

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

$-\frac{51}{50}\int\frac{1}{2u}du+\frac{7}{50}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
24

Take the constant out of the integral

$-\frac{51}{50}\cdot \frac{1}{2}\int\frac{1}{u}du+\frac{7}{50}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
25

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

$\frac{1}{2}-\frac{51}{50}\ln\left|u\right|+\frac{7}{50}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
26

Substitute $u$ back for it's value, $x^2+1$

$\frac{1}{2}-\frac{51}{50}\ln\left|x^2+1\right|+\frac{7}{50}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|$
27

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

$\frac{1}{2}-\frac{51}{50}\ln\left|x^2+1\right|+\frac{7}{50}arctan\left(x\right)+\frac{-\frac{19}{5}}{x-3}+\frac{51}{50}\ln\left|x-3\right|+C_0$

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