12. -/6.25 points details my notes evaluate the integral. (use c for the constant of integration.) ∫…

Answer:

$t\arctan(6t)-\frac{1}{12}\ln(1 + 36t^{2})+C$

Explanation:

Step1: Use integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \arctan(6t)$ and $\mathrm{d}v=\mathrm{d}t$. Then $\mathrm{d}u=\frac{6}{1 + 36t^{2}}\mathrm{d}t$ and $v=t$.

Step2: Apply the formula

$\int\arctan(6t)\mathrm{d}t=t\arctan(6t)-\int\frac{6t}{1 + 36t^{2}}\mathrm{d}t$.

Step3: Solve the new integral

Let $u = 1+36t^{2}$, then $\mathrm{d}u = 72t\mathrm{d}t$. So $\int\frac{6t}{1 + 36t^{2}}\mathrm{d}t=\frac{1}{12}\int\frac{\mathrm{d}u}{u}=\frac{1}{12}\ln|u|+C=\frac{1}{12}\ln(1 + 36t^{2})+C$.

Step4: Get the final result

Combining the results, we have $\int\arctan(6t)\mathrm{d}t=t\arctan(6t)-\frac{1}{12}\ln(1 + 36t^{2})+C$.