# Integral of (7x^3)/((x^4-2)^4)

## \int\frac{7x^3}{\left(x^4-2\right)^4}dx

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$\frac{-\frac{7}{12}}{\left(x^4-2\right)^{3}}+C_0$

## Step by step solution

Problem

$\int\frac{7x^3}{\left(x^4-2\right)^4}dx$
1

Taking the constant out of the integral

$7\int\frac{x^3}{\left(x^4-2\right)^4}dx$
2

Solve the integral $\int\frac{x^3}{\left(x^4-2\right)^4}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x^4-2 \\ du=4x^{3}dx\end{matrix}$
3

Isolate $dx$ in the previous equation

$\frac{du}{4x^{3}}=dx$
4

Substituting $u$ and $dx$ in the integral

$7\int\frac{1}{4u^4}du$
5

Taking the constant out of the integral

$7\cdot \frac{1}{4}\int\frac{1}{u^4}du$
6

Multiply $\frac{1}{4}$ times $7$

$\frac{7}{4}\int\frac{1}{u^4}du$
7

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

$\frac{7}{4}\int u^{-4}du$
8

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

$\frac{7}{4}\cdot\frac{u^{-3}}{-3}$
9

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

$\frac{7}{4}\cdot\frac{\left(x^4-2\right)^{-3}}{-3}$
10

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

$\frac{7}{4}\cdot\frac{\frac{1}{\left(x^4-2\right)^{3}}}{-3}$
11

Simplify the fraction

$-\frac{7}{12}\cdot\frac{1}{\left(x^4-2\right)^{3}}$
12

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=-\frac{7}{12}$ and $x=\left(x^4-2\right)^{3}$

$\frac{-\frac{7}{12}}{\left(x^4-2\right)^{3}}$
13

$\frac{-\frac{7}{12}}{\left(x^4-2\right)^{3}}+C_0$

$\frac{-\frac{7}{12}}{\left(x^4-2\right)^{3}}+C_0$

### Main topic:

Integration by substitution

0.38 seconds

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