x^{-11} (1+x^4)^{-1/2} dx =

Correct Answer is : None of the above

\; -1/2 ≤ft[ \; {(1+x^{-4})^{5/2}}{5} - \; 2/3 (1+x^{-4})³/2} + (1+x^{-4}) ] + c

\; -1/2 ≤ft[ \; {(1+x^4)^{5/2}}{5} - \; 2/3 (1+x^4)³/2} + (1+x^4) ] + c

\; 1/2 ≤ft[ \; {(1+x^4)^{5/2}}{5} + \; 2/3 (1+x^4)³/2} + (1+x^4) ] + c

Correct Answer is : None of the above

Test Index

CGPET 2022 Solved Paper

Section: Mathematics

Question : 136 of 150

Marks: +1, -0

\int x^{-11} (1+x^4)^{-\frac{1}{2}} dx =

$\; -\frac{1}{2} \left[ \; \frac{(1+x^{-4})^{5/2}}{5} - \; \frac{2}{3} (1+x^{-4})^{3/2} + (1+x^{-4}) \right] + c$
$\; -\frac{1}{2} \left[ \; \frac{(1+x^4)^{5/2}}{5} - \; \frac{2}{3} (1+x^4)^{3/2} + (1+x^4) \right] + c$
$\; \frac{1}{2} \left[ \; \frac{(1+x^4)^{5/2}}{5} + \; \frac{2}{3} (1+x^4)^{3/2} + (1+x^4) \right] + c$
None of the above

Solution:

Let

I=\int x^{-11} (1+x^4)^{-\frac{1}{2}} dx

=\int \frac{1}{x^{11} (1+x^4)^{\frac{1}{2}}} dx

Put

x^2=t

\Rightarrow 2x dx = dt

\therefore I=\frac{1}{2} \int \frac{dt}{t^6 (1+t^2)^{\frac{1}{2}}}

Put

t=\tan u

\Rightarrow dt=\sec^2 u du

\therefore I=\frac{1}{2} \int \frac{\sec^2 u du}{\tan^6 u \cdot \sec u}

=\frac{1}{2} \int \frac{\sec u}{\tan^6 u} du

=\frac{1}{2} \int \frac{\cos^5 u}{\sin^6 u} du

=\frac{1}{2} \int \cot^5 u \csc u du

=\frac{1}{2} \int \cot^4 u \csc u \cdot \cot u du

=\frac{1}{2} \int (\csc^2 u-1)^2 \csc u \cot u du

Put

\csc u=v

\Rightarrow -\csc u \cot u du = dv

\therefore I=-\frac{1}{2} \int (v^2-1)^2 dv

=-\frac{1}{2} \int (v^4-2v^2+1) dv

=-\frac{1}{2} \left[ \frac{v^5}{5} - \frac{2v^3}{3} + v \right] + C

=-\frac{1}{2} \left[ \frac{\csc^5 u}{5} - \frac{2}{3} \csc^3 u + \csc u \right] + c

=-\frac{1}{2} \left[ \frac{(t^2+1)^{5/2}}{5 t^5} - \frac{2}{3} \frac{(t^2+1)^{3/2}}{t^3} + \frac{(t^2+1)^{1/2}}{t} \right] + c

=-\frac{1}{2} \left[ \frac{(x^4+1)^{5/2}}{5 x^{10}} - \frac{2}{3} \frac{(x^4+1)^{3/2}}{x^6} + \frac{(x^4+1)^{1/2}}{x^2} \right] + c

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