x^9 (4x^2+1)^6 dx is equal to: - Vidyadip

MCQ

$\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}$ is equal to:

A
$\frac{1}{5\text{x}}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
B
$\frac{1}{5}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
C
$\frac{1}{10\text{x}}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$
✓
$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

✓

Answer

Correct option: D.

$\frac{1}{10}\Big(\frac{1}{\text{x}^2}+4\Big)^{-5}+\text{C}$

Let $\text{I}=\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}$
$=\int\frac{\text{x}^{9}}{\text{x}^{\frac{1}{2}}\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$
$=\int\frac{\text{x}^{\frac{1}{3}}}{\big(4+\frac{1}{\text{x}^2}\big)^6}\text{ dx}$
Let $\Big(4+\frac{1}{\text{x}^2}\Big)=\text{t}$
On differentiating both sides, we get
$-\frac{2}{\text{x}^3}\text{dx}=\text{dt}$
$\therefore\ \text{I}=-\frac{1}{2}\int\frac{1}{(\text{t}^{6})}\text{dt}$
$=-\frac{1}{2}\Big(-\frac{1}{5}\Big)\text{t}^{-5}+\text{C}$
$=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
Therefore, $\int\frac{\text{x}^9}{(4\text{x}^2+1)^6}\text{ dx}=\frac{1}{10}\Big(4+\frac{1}{\text{x}^2}\Big)^{-5}+\text{C}$
Hence, the correct option is $(d)$

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