Integrate <mrow> 5 x 3 </mrow> </

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

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

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

Apply basic rules of exponents.

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

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

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

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

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

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

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

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}$.

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

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

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

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

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

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

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

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

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

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}$.

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

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

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

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

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

Simplify.

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

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

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

Simplify.

${\displaystyle \frac{5x^4}{8}+\frac{1}{2x}+\frac{3x^2}{4}+\frac{\sin \left(u_1\right)}{2}-\frac{2}{3}\sin \left(u_2\right)+C}$

Substitute back in for each integration substitution variable.

${\displaystyle \frac{5x^4}{8}+\frac{1}{2x}+\frac{3x^2}{4}+\frac{\sin \left(2x\right)}{2}-\frac{2}{3}\sin \left(3x\right)+C}$

Reorder terms.

$\frac{5}{8}{x}^4+\frac{1}{2x}+\frac{1}{4}3x^2+\frac{1}{2}\sin \left(2x\right)-\frac{2}{3}\sin \left(3x\right)+C$