Integral of (2x+7)^3

\int\left(2x+7\right)^3dx

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$\frac{1}{8}\left(7+2x\right)^{4}+C_0$

Step by step solution

Problem

$\int\left(2x+7\right)^3dx$
1

Solve the integral $\int\left(7+2x\right)^3dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=7+2x \\ du=2dx\end{matrix}$
2

Isolate $dx$ in the previous equation

$\frac{du}{2}=dx$
3

Substituting $u$ and $dx$ in the integral

$\int\frac{u^3}{2}du$
4

Taking the constant out of the integral

$\frac{1}{2}\int u^3du$
5

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

$\frac{1}{2}\cdot\frac{u^{4}}{4}$
6

Substitute $u$ back for it's value, $7+2x$

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

Simplify the fraction

$\frac{1}{8}\left(7+2x\right)^{4}$
8

$\frac{1}{8}\left(7+2x\right)^{4}+C_0$

$\frac{1}{8}\left(7+2x\right)^{4}+C_0$

Main topic:

Integration by substitution

0.25 seconds

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