Evaluate 1 x 4 </mrow> − 1 <mro

Remove parentheses.

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

Split the single integral into multiple integrals.

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

Apply basic rules of exponents.

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

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

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

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

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

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

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

Apply basic rules of exponents.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify.

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

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

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

Simplify.

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

Replace all occurrences of ${\displaystyle u}$ with ${\displaystyle 2x}$.

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

Reorder terms.

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