Question

$\int \frac{x^9}{{\left(4x^2+1\right)}^6}\mathrm{d}x$ is equal to

$$\begin{gathered} \left(\mathrm{a}\right)\frac{1}{5x}{\left(4+\frac{1}{x^2}\right)}^{-5}+C \\ \left(\mathrm{b}\right)\frac{1}{5}{\left(4+\frac{1}{x^2}\right)}^{-5}+C \\ \left(\mathrm{c}\right)\frac{1}{10x}{\left(\frac{1}{x^2}+4\right)}^{-5}+C \\ \left(\mathrm{d}\right)\frac{1}{10}{\left(\frac{1}{x^2}+4\right)}^{-5}+C \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^9}{{\left(4x^2+1\right)}^6}\mathrm{d}x \\ =\int \frac{x^9}{x^{12}{\left(4+{\displaystyle \frac{1}{x^2}}\right)}^6}\mathrm{d}x \\ =\int \frac{{\displaystyle \frac{1}{x^3}}}{{\left(4+{\displaystyle \frac{1}{x^2}}\right)}^6}\mathrm{d}x \\ \mathrm{Let}\left(4+\frac{1}{x^2}\right)=t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ -\frac{2}{x^3}\mathrm{d}x=\mathrm{d}t \\ \therefore I=-\frac{1}{2}\int \frac{1}{{\left(t\right)}^6}\mathrm{d}t \\ =-\frac{1}{2}\left(-\frac{1}{5}\right)t^{-5}+C \\ =\frac{1}{10}{\left(4+\frac{1}{x^2}\right)}^{-5}+C \\ \mathrm{Therefore},\int \frac{x^9}{{\left(4x^2+1\right)}^6}\mathrm{d}x=\frac{1}{10}{\left(4+\frac{1}{x^2}\right)}^{-5}+C \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{d}\right). \end{gathered}$$