Evaluating $\int \arctan4t \, dt$ I am trying to evaluate $$\int \arctan4t \, dt$$
I make $u = \arctan4t$, $du = \frac{4}{1-4t}\,dt$
and $dv = dt$ and $v = t$
I then make it the form of $uv = \int v \, du$
$t\arctan4t  - \int \frac{4t}{1-16t^2} \, dt$
Now to get the integral of $\int \frac{4t}{1-16t^2} \, dt$ I just pull out the 4 and use the identity I have memorized.
But I just now realised that this is not possible since I have a $16t^2$ so I have no idea how to advance from here.
 A: Since $\frac{\mathrm{d}}{\mathrm{d}t}\arctan(4t)=\frac{4}{1+16t^2}$, integration by parts yields
$$
\int\color{red}{\arctan(4t)}\,\color{green}{\mathrm{d}t}=\color{green}{t}\color{red}{\arctan(4t)}-\int\color{green}{t}\color{red}{\frac{4}{1+16t^2}\mathrm{d}t}
$$
Then let $u=1+16t^2$, so that $\frac{\mathrm{d}}{\mathrm{d}t}u=32t$, and we get
$$
\begin{align}
\int\arctan(4t)\,\mathrm{d}t
&=t\arctan(4t)-\int t\frac{4}{1+16t^2}\mathrm{d}t\\
&=t\arctan(4t)-\frac18\int\frac{\mathrm{d}u}{u}\\
&=t\arctan(4t)-\frac18\log(u)+C\\
&=t\arctan(4t)-\frac18\log(1+16t^2)+C
\end{align}
$$
A: Let $s = 1+16t^2$. So $ds = 32t \;dt$ and thus $t \; dt = ds/32$. So we have that 
$$
\int \dfrac{4t}{1+16t^2}dt = \dfrac{1}{8} \int \dfrac{ds}{s} = \dfrac{1}{8} \mbox{ln $|s|$} + C = \dfrac{1}{8} \mbox{ln $|1+16t^2|$} + C.
$$
A: As noted by David in the comments, your $du$ is off. Your $du$ should be $\frac{4 \, dt}{1+16t^2}.$
So, we have:
$$t\arctan(4t)  - \int \frac{4t \, dt}{1+16t^2}$$
Then, you can let $s = 1 + 16t^2$ and $ds = 32t \, dt$.
$$t\arctan(4t)  - \frac{1}{8} \int \frac{1}{s} \, ds$$
$$t\arctan(4t)  - \frac{1}{8} \ln {(1+16t^2)} + C$$
A: More generally
$$\int \arctan x dx=x\arctan x-\frac{1}{2} \log\left( 1+x^2\right)+C$$
So that with $au=x$
$$\int \arctan au du =\frac{1}{a}\int\arctan x dx=\frac{x}{a}\arctan x-\frac{1}{2a} \log\left( 1+x^2\right)+C$$
$$\int \arctan au du ={u}\arctan au-\frac{1}{2a} \log\left( 1+a^2u^2\right)+C$$
The first result is obtain quite striaghtforwarly from integrating by parts with $dx=du$ and $\arctan x =v$. 
$$\eqalign{
  & \int {\arctan } xdx = x\arctan x - \int {\frac{{xdx}}{{1 + {x^2}}}}   \cr 
  & \int {\arctan } xdx = x\arctan x - \frac{1}{2}\int {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}}   \cr 
  & \int {\arctan } xdx = x\arctan x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + C \cr} $$
