# $\Large \int\frac{x^9}{(4x^2+1)^6}\normalsize dx$ is equal to\begin{array}{1 1}(A)\;\frac{1}{5x}\bigg(4+\frac{1}{x^2}\bigg)^{-5}+C & (B)\;\frac{1}{5}\bigg(4+\frac{1}{x^2}\bigg)^{-5}+C\\(C)\;\frac{1}{10x}(1+4)^{-5}+C & (D)\;\frac{1}{10}\bigg(\frac{1}{x^2}+4\bigg)^{-5}+C\end{array}

Toolbox:
• If $f(x)$ is substituted by $f(t)$,then $f'(x)dx=f'(t)dt.$
• Hence $\int f(x)dx=\int f(t)dt.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c Step 1: Let I=\int \large\frac{x^9}{(4x^2+1)^6}$$dx$
$4x^2+1$ can be written as $x^2(4+\large\frac{1}{x^2}).$
$\Rightarrow \begin{bmatrix}x^2(4+\large\frac{1}{x^2})\end{bmatrix}^6$=$x^{12}(4+\large\frac{1}{x^2})^6$
$I=\int \large\frac{x^9}{x^{12}(4+\large\frac{1}{x^2})^6}$$dx \;\;=\int \large\frac{dx}{x^3(4+\large\frac{1}{x^2})^6}$$dx$
Step 2:
Put $(4+\large\frac{1}{x^2})$=$t$.
On differentiating with respect to t we get,
$\large\frac{-2}{x^3}$$dx=dt\Rightarrow\large\frac{dx}{x^3}=\frac{-dt}{2} Now substituting for this we get, I=\int \large\frac{-dt/2}{t^6}=\frac{-1}{2}\int t^{-6}$$dt$
On integrating we get,
$\large\frac{-1}{2}\begin{bmatrix}\large\frac{t^{-6+1}}{-6+1}\end{bmatrix}=\frac{-1}{2}\frac{-1}{5}\begin{bmatrix}\frac{1}{t^5}\end{bmatrix}=\frac{1}{10}\big(\frac{1}{t^5}\big)$
Substituting for $t$ we get,
$I=\large\frac{1}{10}\begin{bmatrix}\frac{1}{4+\large\frac{1}{x^2}}\end{bmatrix}$$+c=\frac{1}{10}\big(\large\frac{1}{x^2}+\normalsize 4\big)^{-5}+c.$
Hence D is the correct option.