Evaluate 3 <mrow> 2 x 4 </mrow>

Remove parentheses.

${\displaystyle \int \frac{3}{2x^4}-\frac{2}{x}+\frac{3}{2}\cdot \sin \left(2x\right)+\frac{5}{2}\cdot \cos \left(3x\right)dx}$

Split the single integral into multiple integrals.

${\displaystyle \int \frac{3}{2x^4}dx+\int -\frac{2}{x}dx+\int \frac{3}{2}\cdot \sin \left(2x\right)dx+\int \frac{5}{2}\cdot \cos \left(3x\right)dx}$

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

${\displaystyle \frac{3}{2}\int \frac{1}{x^4}dx+\int -\frac{2}{x}dx+\int \frac{3}{2}\cdot \sin \left(2x\right)dx+\int \frac{5}{2}\cdot \cos \left(3x\right)dx}$

Apply basic rules of exponents.

${\displaystyle \frac{3}{2}\int x^{-4}dx+\int -\frac{2}{x}dx+\int \frac{3}{2}\cdot \sin \left(2x\right)dx+\int \frac{5}{2}\cdot \cos \left(3x\right)dx}$

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

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

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

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

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

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

Multiply ${\displaystyle 2}$ by ${\displaystyle -1}$.

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

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

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

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

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

Let ${\displaystyle u_1=2x}$. Then ${\displaystyle d{u}_1=2dx}$, so ${\displaystyle \frac{1}{2}d{u}_1=dx}$. Rewrite using ${\displaystyle u_1}$ and ${\displaystyle d}$${\displaystyle u_1}$.

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

Combine ${\displaystyle \sin \left(u_1\right)}$ and ${\displaystyle \frac{1}{2}}$.

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

Since ${\displaystyle \frac{1}{2}}$ is constant with respect to ${\displaystyle u_1}$, move ${\displaystyle \frac{1}{2}}$ out of the integral.

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

Simplify.

${\displaystyle \frac{3}{2}(-\frac{1}{3}{x}^{-3}+C)-2(\ln \left(\left|x\right|\right)+C)+\frac{3}{4}\int \sin \left(u_1\right)d{u}_1+\int \frac{5}{2}\cdot \cos \left(3x\right)dx}$

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

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

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

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

Let ${\displaystyle u_2=3x}$. Then ${\displaystyle d{u}_2=3dx}$, so ${\displaystyle \frac{1}{3}d{u}_2=dx}$. Rewrite using ${\displaystyle u_2}$ and ${\displaystyle d}$${\displaystyle u_2}$.

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

Combine ${\displaystyle \cos \left(u_2\right)}$ and ${\displaystyle \frac{1}{3}}$.

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

Since ${\displaystyle \frac{1}{3}}$ is constant with respect to ${\displaystyle u_2}$, move ${\displaystyle \frac{1}{3}}$ out of the integral.

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

Simplify.

${\displaystyle \frac{3}{2}(-\frac{1}{3}{x}^{-3}+C)-2(\ln \left(\left|x\right|\right)+C)+\frac{3}{4}(-\cos \left(u_1\right)+C)+\frac{5}{6}\int \cos \left(u_2\right)d{u}_2}$

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

${\displaystyle \frac{3}{2}(-\frac{1}{3}{x}^{-3}+C)-2(\ln \left(\left|x\right|\right)+C)+\frac{3}{4}(-\cos \left(u_1\right)+C)+\frac{5}{6}(\sin \left(u_2\right)+C)}$

Simplify.

${\displaystyle -\frac{1}{2x^3}-2\ln \left(\left|x\right|\right)-\frac{3\cos \left(u_1\right)}{4}+\frac{5}{6}\sin \left(u_2\right)+C}$

Substitute back in for each integration substitution variable.

${\displaystyle -\frac{1}{2x^3}-2\ln \left(\left|x\right|\right)-\frac{3\cos \left(2x\right)}{4}+\frac{5}{6}\sin \left(3x\right)+C}$

Reorder terms.

$-\frac{1}{2x^3}-2\ln \left(\left|x\right|\right)-\frac{3}{4}\cos \left(2x\right)+\frac{5}{6}\sin \left(3x\right)+C$