use the table of integrals to evaluate the following integral. \int 3x \sqrt{6x-x^2}dx

Evaluate the integral.
${\displaystyle \int \frac{\cos (1-\ln \left(y\right))}{y}dy}$

${\displaystyle \int \frac{dx}{1+x},x\ge 0}$ by trigonometric substitution
Consider the integral ${\displaystyle I=\int \frac{dx}{1+x},x\in [0,\mathrm{\infty })}$. A standard treatment using the substitution u=1+x directly gives the result ${\displaystyle \ln (1+x)+c}$. Now consider doing this with the trigonometric substitution ${\displaystyle \sqrt{x}=\tan \theta ,\theta \in [0,\frac{\pi }{2})}$ then ${\displaystyle x={\tan }^2\theta ,dx=2\tan \theta {\sec }^2\theta }$. Now, following your nose with right triangle trigonometry this leads directly to the solution ${\displaystyle I=2\ln (1+x)+c}$.

Evaluate the following integrals. Include absolute values.
${\displaystyle {\int }_0^1\frac{y{\ln }^4(y^2+1)}{y^2+1}dy}$

How do you find the primitive/integral of ${\displaystyle \arctan x}$?
I tried searching for how you derive the integral/primitive of ${\displaystyle \arctan \left(x\right)}$, but I can't find any question on S.E with an answer that clearly explains this. I feel that there should be one, since it is rather fundamental knowledge, and it does come up every now and then as part of S.E questions.
So the primitive can be looked up and it should be the following:
${\displaystyle \int \arctan \left(x\right)dx=x\arctan \left(x\right)-\frac{1}{2}\ln (1+x^2)+C}$
How do you get the result?

Evaluate $\int \frac{4}{9}e^{6x}\tan (e^{6x}+4)\sec (e^{6x}+4)dx$.

Evaluate the indefinite integral.
${\displaystyle \int \frac{x}{\sqrt{x^2+9}}dx}$

Evaluate the integral.
${\displaystyle \int \frac{{(1+e^x)}^2}{e^x}dx}$