∫(x^9)/((4x^2+1)^6) dx is equal to:[where c is an arbitrary constant]

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

[where $ c $ is an arbitrary constant]

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → $ \displaystyle\frac{1}{10}\left(\displaystyle\frac{1}{x^2}+4\right)^{-5}+c $

The denominator has degree 12 and the numerator degree 9. Taking $ x^2 $ out of the bracket makes a function and its derivative appear:

$ \displaystyle (4x^2+1)^6=\left[x^2\left(4+\frac{1}{x^2}\right)\right]^6 $

$ \displaystyle =x^{12}\left(4+\frac{1}{x^2}\right)^6 $

$ \displaystyle \int\frac{x^9}{(4x^2+1)^6}\,dx=\int\frac{1}{x^3\left(4+\frac{1}{x^2}\right)^6}\,dx $

Let $ t=4+\frac{1}{x^2} $ (a substitution). Then

$ \displaystyle dt=-\frac{2}{x^3}\,dx $

$ \displaystyle \frac{dx}{x^3}=-\frac{dt}{2} $

$ \displaystyle \int\frac{1}{x^3\left(4+\frac{1}{x^2}\right)^6}\,dx=-\frac{1}{2}\int t^{-6}\,dt $

$ \displaystyle =-\frac{1}{2}\cdot\frac{t^{-5}}{-5}+c $

$ \displaystyle =\frac{1}{10}\,t^{-5}+c $

$ \displaystyle =\frac{1}{10}\left(\frac{1}{x^2}+4\right)^{-5}+c $

Shortcut: Differentiate Option (4):

$ \displaystyle \frac{d}{dx}\left[\frac{1}{10}\left(x^{-2}+4\right)^{-5}\right]=\frac{1}{10}(-5)\left(x^{-2}+4\right)^{-6}\left(-2x^{-3}\right) $

$ \displaystyle =\frac{1}{x^3\left(4+\frac{1}{x^2}\right)^6} $

Multiplying the top and bottom by $ x^{12} $ gives $ \frac{x^9}{(4x^2+1)^6} $, the integrand.

Why the other options are incorrect

Option (1) $ \displaystyle\frac{1}{5x}\left(4+\displaystyle\frac{1}{x^2}\right)^{-5}+c $ — Incorrect. The factor $ \frac{1}{x} $ cannot be taken outside after integrating; its derivative contains extra terms from $ \frac{1}{x} $.
Option (2) $ \displaystyle\frac{1}{5}\left(4+\displaystyle\frac{1}{x^2}\right)^{-5}+c $ — Incorrect. The factor $ \frac{1}{2} $ from $ dt=-\frac{2}{x^3}dx $ has been missed; this is twice the correct answer.
Option (3) $ \displaystyle\frac{1}{10x}(1+4)^{-5}+c $ — Incorrect. The bracket $ (1+4) $ contains no x, so its derivative is $ -\frac{5^{-5}}{10x^2} $, not the integrand.

NCERT Exemplar:
Integrals

$NCERT extract$