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Remove parentheses.

${\displaystyle \int -\frac{3}{x^3}+\frac{3}{2x}-3\sin \left(3x\right)-\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Split the single integral into multiple integrals.

${\displaystyle \int -\frac{3}{x^3}dx+\int \frac{3}{2x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

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

${\displaystyle -\int \frac{3}{x^3}dx+\int \frac{3}{2x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

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

${\displaystyle -(3\int \frac{1}{x^3}dx)+\int \frac{3}{2x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Simplify the expression.

${\displaystyle -3\int x^{-3}dx+\int \frac{3}{2x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

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

${\displaystyle -3(-\frac{1}{2}{x}^{-2}+C)+\int \frac{3}{2x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Simplify.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\int \frac{3}{2x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

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

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}\int \frac{1}{x}dx+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

The integral of ${\displaystyle \frac{1}{x}}$ with respect to ${\displaystyle x}$ is ${\displaystyle \ln \left(\left|x\right|\right)}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)+\int -3\sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

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

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-3\int \sin \left(3x\right)dx+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Let ${\displaystyle u_1=3x}$. Then ${\displaystyle d{u}_1=3dx}$, so ${\displaystyle \frac{1}{3}d{u}_1=dx}$. Rewrite using ${\displaystyle u_1}$ and ${\displaystyle d}$${\displaystyle u_1}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-3\int \sin \left(u_1\right)\frac{1}{3}d{u}_1+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Combine ${\displaystyle \sin \left(u_1\right)}$ and ${\displaystyle \frac{1}{3}}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-3\int \frac{\sin \left(u_1\right)}{3}d{u}_1+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Since ${\displaystyle \frac{1}{3}}$ is constant with respect to ${\displaystyle u_1}$, move ${\displaystyle \frac{1}{3}}$ out of the integral.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-3(\frac{1}{3}\int \sin \left(u_1\right)d{u}_1)+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Simplify.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-\int \sin \left(u_1\right)d{u}_1+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

The integral of ${\displaystyle \sin \left(u_1\right)}$ with respect to ${\displaystyle u_1}$ is ${\displaystyle -\cos \left(u_1\right)}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)+\int -\frac{5}{2}\cdot \cos \left(2x\right)dx}$

Since ${\displaystyle -\frac{5}{2}}$ is constant with respect to ${\displaystyle x}$, move ${\displaystyle -\frac{5}{2}}$ out of the integral.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)-\frac{5}{2}\int \cos \left(2x\right)dx}$

Let ${\displaystyle u_2=2x}$. Then ${\displaystyle d{u}_2=2dx}$, so ${\displaystyle \frac{1}{2}d{u}_2=dx}$. Rewrite using ${\displaystyle u_2}$ and ${\displaystyle d}$${\displaystyle u_2}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)-\frac{5}{2}\int \cos \left(u_2\right)\frac{1}{2}d{u}_2}$

Combine ${\displaystyle \cos \left(u_2\right)}$ and ${\displaystyle \frac{1}{2}}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)-\frac{5}{2}\int \frac{\cos \left(u_2\right)}{2}d{u}_2}$

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)-\frac{5}{2}(\frac{1}{2}\int \cos \left(u_2\right)d{u}_2)}$

Simplify.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)-\frac{5}{4}\int \cos \left(u_2\right)d{u}_2}$

The integral of ${\displaystyle \cos \left(u_2\right)}$ with respect to ${\displaystyle u_2}$ is ${\displaystyle \sin \left(u_2\right)}$.

${\displaystyle -3(-\frac{1}{2x^2}+C)+\frac{3}{2}(\ln \left(\left|x\right|\right)+C)-(-\cos \left(u_1\right)+C)-\frac{5}{4}(\sin \left(u_2\right)+C)}$

Simplify.

${\displaystyle \frac{3}{2x^2}+\frac{3\ln \left(\left|x\right|\right)}{2}+\cos \left(u_1\right)-\frac{5}{4}\sin \left(u_2\right)+C}$

Substitute back in for each integration substitution variable.

${\displaystyle \frac{3}{2x^2}+\frac{3\ln \left(\left|x\right|\right)}{2}+\cos \left(3x\right)-\frac{5}{4}\sin \left(2x\right)+C}$

Reorder terms.

$\frac{3}{2x^2}+\frac{3}{2}\ln \left(\left|x\right|\right)+\cos \left(3x\right)-\frac{5}{4}\sin \left(2x\right)+C$