Question

Evaluate:$\frac{1}{15}\int \frac{dt}{4+t^2}+\frac{4}{15}\int \frac{6dt}{1+4t^2}$

Answer 1

$\frac{1}{15}\int \frac{dt}{4+t^2}+\frac{4}{15}\int \frac{6dt}{1+4t^2}$

$=\frac{1}{15}\cdot \frac{1}{2}{\tan }^{-1}\left(\frac{t}{2}\right)+\frac{24}{15\times 4}\int \frac{dt}{{\left(\frac{1}{2}\right)}^2+t^2}$

$=\frac{1}{30}{\tan }^{-1}\left(\frac{t}{2}\right)+\frac{6}{15}\cdot \frac{1}{1/2}{\tan }^{-1}\left(\frac{t}{\frac{1}{2}}\right)+c$

$=\frac{1}{30}{\tan }^{-1}\left(\frac{t}{2}\right)+\frac{4}{5}{\tan }^{-1}\left(2t\right)+c$.