Question

Evaluate : $I=\int \frac{x+4}{x^2+5}dx$

Answer 1

$I=\int \left(\frac{x+4}{x^2+5}\right)dx$

$=\int \left(\frac{x}{x^2+5}\right)dx+4\int \left(\frac{1}{x^2+5}\right)dx$

$Use{x}^2+5=t,then2xdx=dt$

$anduseformula=\int \left(\frac{1}{a^2+x^2}\right)dx=\left(\frac{1}{a}\right)ta{n}^{-1}\left(\frac{x}{a}\right)$d

$\therefore I=\left(\frac{1}{2}\right)\int \left(\frac{dt}{t}\right)+4\int \left(\frac{1}{x^2+(\sqrt{5}{)}^2}\right)dx$

$=\left(\frac{1}{2}\right)logt+4\cdot \left(\frac{1}{\sqrt{5}}\right)ta{n}^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$

$=\left(\frac{1}{2}\right)log(x^2+5)+\left(\frac{4}{\sqrt{5}}\right)ta{n}^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$