Solving the Integral using $\ln|u|$ Can Someone help me solve this
$$
\int\frac{19\tan^{-1}x}{x^{2}}\,dx
$$
We have been told to use $\ln|u|$ and $C$.
Thanks!
 A: Using integration by parts we see:
$$ \int \frac{ \arctan(x) }{x^2} dx = -\frac{\arctan(x) }{x} + \int \frac{dx}{ (x^2+1) x} $$
Now if we use a partial fraction decomposition we obtain:
$$\int \frac{dx}{ (x^2+1) x} = \int \left (  \frac{1}{x} - \frac{x}{x^2 +1} \right ) dx $$
From here it should be clear that
$$\int \frac{ \arctan(x) }{x^2} dx = \ln (x) - \frac{1}{2} \ln (x^2 +1) - \frac{\arctan(x)}{x} +C $$
A: Let $y=\arctan x\;\Rightarrow\;\tan y=x\;\Rightarrow\;\sec^2y\ dy=dx$, then the integral turns out to be
\begin{align}
\int\frac{\arctan x}{x^2}\ dx&=\int\frac{y}{\tan^2y}\cdot\sec^2y\ dy\\
&=\int\frac{y}{\sin^2y}\ dy.
\end{align}
The last integral can be solved by using IBP. Taking $u=y$ and $dv=\dfrac1{\sin^2x}\ dx\;\Rightarrow\;v=-\cot y$.
\begin{align}
\int\frac{\arctan x}{x^2}\ dx&=\int\frac{y}{\sin^2y}\ dy\\
&=-y\cot y+\int\cot y\ dy\\
&=-y\cot y+\int\frac{\cos y}{\sin y}\ dy\\
&=-y\cot y+\int\frac{d(\sin y)}{\sin y}\\
&=-y\cot y+\ln|\sin y|+C\\
&=-\frac{\arctan x}{x}+\ln\left|\frac{x}{\sqrt{x^2+1}}\right|+C.
\end{align}
