Question

Solve $\int \left[\frac{1}{t^2+3+2t}\right]dt$

Answer 1

Formula:

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+c$

$\int \left[\frac{1}{t^2+2t+3}\right]dt=\int \left[\frac{1}{(t+1{)}^2+2}\right]dt$

$=\int \frac{1}{(t+1{)}^2+(\sqrt{2}{)}^2}dt$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+c$