Question

Integral using partial fraction${\int }_1^2\frac{dx}{x+x^3}=$

Answer 1

${\int }_1^2\frac{dx}{x+x^3}=\int \frac{1}{x^3}\left(\frac{dx}{\frac{1}{x^2}+1}\right)$

$\Rightarrow \frac{1}{x^2}+1=t$

$\Rightarrow \frac{-2}{x^3}=\frac{dt}{dx}\Rightarrow dx=\frac{-x^3}{2}dt$

$\Rightarrow \frac{-1}{2}\int \left(\frac{1}{t}\right)dt=\frac{-1}{2}\ln \left(t\right)+C$

$={\left[\frac{-1}{2}\ln \left(\frac{1+x^2}{x^2}\right)\right]}_1^{2}t$

$\Rightarrow \frac{-1}{2}[\ln \left(\frac{5}{4}\right)-\ln (2\left)\right]$

$\Rightarrow -\frac{1}{2}\ln \left(\frac{5}{8}\right)$