Question

evaluate the integral using integration by parts.
int\tan^{-1}(2y)dy
int\tan^{-1}(2y)dy=square
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \tan^{- 1}(2y)$ and $\mathrm{d}v=\mathrm{d}y$. Then $\mathrm{d}u=\frac{2}{1 + 4y^{2}}\mathrm{d}y$ and $v = y$.

Step2: Substitute into the formula

$\int\tan^{-1}(2y)\mathrm{d}y=y\tan^{-1}(2y)-\int y\cdot\frac{2}{1 + 4y^{2}}\mathrm{d}y$.

Step3: Evaluate the new integral

Let $t = 1+4y^{2}$, then $\mathrm{d}t = 8y\mathrm{d}y$ and $y\mathrm{d}y=\frac{1}{8}\mathrm{d}t$. So $\int\frac{2y}{1 + 4y^{2}}\mathrm{d}y=\frac{2}{8}\int\frac{\mathrm{d}t}{t}=\frac{1}{4}\ln|t|+C=\frac{1}{4}\ln(1 + 4y^{2})+C$.

Answer:

$y\tan^{-1}(2y)-\frac{1}{4}\ln(1 + 4y^{2})+C$