Compute x 3 </mrow> 2 − 2

Remove parentheses.

${\displaystyle \int \frac{x^3}{2}-\frac{2}{x^3}+\frac{3\sin \left(x\right)}{2}+2\cos \left(3x\right)dx}$

Split the single integral into multiple integrals.

${\displaystyle \int \frac{x^3}{2}dx+\int -\frac{2}{x^3}dx+\int \frac{3\sin \left(x\right)}{2}dx+\int 2\cos \left(3x\right)dx}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

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

By the Power Rule, the integral of ${\displaystyle x^3}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{1}{4}{x}^4}$.

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

Since ${\displaystyle -1}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle -1}$ out of the integral.

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

Since ${\displaystyle 2}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle 2}$ out of the integral.

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

Simplify the expression.

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

By the Power Rule, the integral of ${\displaystyle x^{-3}}$ with respect to ${\displaystyle x}$ is ${\displaystyle -\frac{1}{2}{x}^{-2}}$.

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

Simplify.

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

Since ${\displaystyle \frac{3}{2}}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle \frac{3}{2}}$ out of the integral.

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

The integral of ${\displaystyle \sin \left(x\right)}$ with respect to ${\displaystyle x}$ is ${\displaystyle -\cos \left(x\right)}$.

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

Since ${\displaystyle 2}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle 2}$ out of the integral.

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

Let ${\displaystyle u=3x}$. Then ${\displaystyle du=3dx}$, so ${\displaystyle \frac{1}{3}du=dx}$. Rewrite using ${\displaystyle u}$ and ${\displaystyle d}$${\displaystyle u}$.

${\displaystyle \frac{1}{2}(\frac{1}{4}{x}^4+C)-2(-\frac{1}{2x^2}+C)+\frac{3}{2}(-\cos \left(x\right)+C)+2\int \cos \left(u\right)\frac{1}{3}du}$

Combine ${\displaystyle \cos \left(u\right)}$ and ${\displaystyle \frac{1}{3}}$.

${\displaystyle \frac{1}{2}(\frac{1}{4}{x}^4+C)-2(-\frac{1}{2x^2}+C)+\frac{3}{2}(-\cos \left(x\right)+C)+2\int \frac{\cos \left(u\right)}{3}du}$

Since ${\displaystyle \frac{1}{3}}$ is constant with respect to ${\displaystyle u}$, move ${\displaystyle \frac{1}{3}}$ out of the integral.

${\displaystyle \frac{1}{2}(\frac{1}{4}{x}^4+C)-2(-\frac{1}{2x^2}+C)+\frac{3}{2}(-\cos \left(x\right)+C)+2(\frac{1}{3}\int \cos \left(u\right)du)}$

Combine ${\displaystyle \frac{1}{3}}$ and ${\displaystyle 2}$.

${\displaystyle \frac{1}{2}(\frac{1}{4}{x}^4+C)-2(-\frac{1}{2x^2}+C)+\frac{3}{2}(-\cos \left(x\right)+C)+\frac{2}{3}\int \cos \left(u\right)du}$

The integral of ${\displaystyle \cos \left(u\right)}$ with respect to ${\displaystyle u}$ is ${\displaystyle \sin \left(u\right)}$.

${\displaystyle \frac{1}{2}(\frac{1}{4}{x}^4+C)-2(-\frac{1}{2x^2}+C)+\frac{3}{2}(-\cos \left(x\right)+C)+\frac{2}{3}(\sin \left(u\right)+C)}$

Simplify.

${\displaystyle \frac{x^4}{8}+\frac{1}{x^2}-\frac{3\cos \left(x\right)}{2}+\frac{2}{3}\sin \left(u\right)+C}$

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle 3x}$.

${\displaystyle \frac{x^4}{8}+\frac{1}{x^2}-\frac{3\cos \left(x\right)}{2}+\frac{2}{3}\sin \left(3x\right)+C}$

Reorder terms.

$\frac{1}{8}{x}^4+\frac{1}{x^2}-\frac{3}{2}\cos \left(x\right)+\frac{2}{3}\sin \left(3x\right)+C$