Integral of x(x^2-3)^3*8

\int8x\left(x^2-3\right)^3dx

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Answer

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

Step by step solution

Problem

$\int8x\left(x^2-3\right)^3dx$
1

Taking the constant out of the integral

$8\int x\left(x^2-3\right)^3dx$
2

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

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

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
4

Substituting $u$ and $dx$ in the integral

$8\int\frac{u^3}{2}du$
5

Taking the constant out of the integral

$8\cdot \frac{1}{2}\int u^3du$
6

Multiply $\frac{1}{2}$ times $8$

$4\int u^3du$
7

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

$4\frac{u^{4}}{4}$
8

Substitute $u$ back for it's value, $x^2-3$

$4\frac{\left(x^2-3\right)^{4}}{4}$
9

Simplify the fraction

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

Add the constant of integration

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

Answer

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

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.29 seconds

Views:

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