Originally Posted by

**Chris L T521** The terms in red should be 2, not 4.

Now, from here, do the following:

We found that $\displaystyle \int 5y\tan^{-1}(5y)\,dy=$ $\displaystyle 10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}-\frac {1}{2}{\color{red}\int} \left({\color{red}5y\tan^{-1}(5y)}-\tfrac {1}{2}\ln{(1+25y^2)}\right){\color{red}\,dy}$

The integral reappears on the right side! So get it on the left side!

$\displaystyle \implies \tfrac{3}{2}\int 5y\tan^{-1}(5y)\,dy=10y^2\tan^{-1}(5y)-\tfrac {1}{2}\ln{(1+25y^2)}+\tfrac{1}{4}{\color{red}\int\ ln{(1+25y^2)}\,dy}$

Before we get our answer, try to evaluate the integral I highlighted in red.

Hint : apply integration by parts! Let $\displaystyle u=\ln(1+25y^2)$.....