Evaluate 2 x 3 </mrow> − 3 <mrow> 2 x

Remove parentheses.

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

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

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

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

Combine ${\displaystyle \frac{1}{4}}$ and ${\displaystyle x^4}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify.

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

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

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

Apply the constant rule.

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

Simplify.

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

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

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

Reorder terms.

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