Question

$\int \frac{x^2}{\left(x^2+1\right)\left(3x^2+4\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{x^{2}dx}{\left(x^2+1\right)\left(3x^2+4\right)} \\ \mathrm{Putting}{x}^2=t \\ \mathrm{Then},\frac{x^2}{\left(x^2+1\right)\left(3x^2+4\right)}=\frac{t}{\left(t+1\right)\left(3t+4\right)} \\ \mathrm{Let}\frac{t}{\left(t+1\right)\left(3t+4\right)}=\frac{A}{t+1}+\frac{B}{3t+4} \\ \Rightarrow \frac{t}{\left(t+1\right)\left(3t+4\right)}=\frac{A\left(3t+4\right)+B\left(t+1\right)}{\left(t+1\right)\left(3t+4\right)} \\ \Rightarrow t=A\left(3t+4\right)+B\left(t+1\right) \\ \mathrm{Putting}t+1=0 \\ \Rightarrow t=-1 \\ \therefore -1=A\left(-3+4\right)+0 \\ \Rightarrow A=-1 \\ \mathrm{Putting}3t+4=0 \\ \Rightarrow t=-\frac{4}{3} \\ \therefore -\frac{4}{3}=0+B\left(-\frac{4}{3}+1\right) \\ \Rightarrow -\frac{4}{3}=B\times \left(-\frac{1}{3}\right) \\ \Rightarrow B=4 \\ \therefore \frac{t}{\left(t+1\right)\left(3t+4\right)}=-\frac{1}{t+1}+\frac{4}{3t+4} \\ \Rightarrow \frac{x^2}{\left(x^2+1\right)\left(3x^2+4\right)}=\frac{-1}{x^2+1}+\frac{4}{3x^2+4} \\ \Rightarrow \frac{x^2}{\left(x^2+1\right)\left(3x^2+4\right)}=\frac{-1}{x^2+1}+\frac{4}{3\left(x^2+{\displaystyle \frac{4}{3}}\right)} \\ \Rightarrow \int \frac{x^{2}dx}{\left(x^2+1\right)\left(3x^2+4\right)}=-\int \frac{dx}{x^2+1}+\frac{4}{3}\int \frac{dx}{x^2+{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2} \\ =-{\tan }^{-1}\left(x\right)+\frac{4}{3}\times \frac{\sqrt{3}}{2}{\tan }^{-1}\left(\frac{\sqrt{3}x}{2}\right)+C \\ =-{\tan }^{-1}\left(x\right)+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}x}{2}\right)+C \end{gathered}$$