Integrate − 3 t 3 </mrow> &#xFEFF; </mrow> + 2

Split the single integral into multiple integrals.

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

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

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

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

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

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

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

Simplify the expression.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify.

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

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

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

Simplify.

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

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

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

Reorder terms.

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