#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 32 Maths Textbook Solution.

$\frac{1}{10}\left[4+\frac{1}{x^{2}}\right]^{-5}+C$

Given:

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

Hint:

Using $\int x^{n} d x$.

Explanation:

Let $\mathrm{I}=\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x$

$=\int \frac{x^{9}}{x^{12}\left(4+\frac{1}{x^{2}}\right)^{6}} d x \ldots \ldots \ldots \text { taking } x^{2} \text { common from denominator }$

\begin{aligned} &=\int \frac{1}{x^{3}\left(4+\frac{1}{x^{2}}\right)^{6}} d x \\ &=\int \frac{\frac{1}{x^{3}}}{\left(4+\frac{1}{x^{2}}\right)^{6}} d x \end{aligned}                                                            $\text { [ Put } \left.4+\frac{1}{x^{2}}=t \Rightarrow \frac{-2}{x^{3}} d x=d t\right]$

\begin{aligned} &=-\frac{1}{2} \int \frac{d t}{t^{6}} \\ &=-\frac{1}{2}\left[\frac{t^{-6+1}}{-6+1}\right]+C \\ &=-\frac{1}{2}\left[\frac{(t)^{-5}}{-5}\right]+C \\ &=\frac{1}{10} \times \frac{1}{t^{5}}+C \\ &=\frac{t^{-5}}{10}+C \\ &=\frac{1}{10}\left[4+\frac{1}{x^{2}}\right]^{-5}+C \end{aligned}

Hence,$\int \frac{x^{9}}{\left(4 x^{2}+1\right)^{6}} d x=\frac{1}{10}\left[4+\frac{1}{x^{2}}\right]^{-5}+C$