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Question: Evaluate \(\int_{}^{}{x^{- 11}(1 + x^{4})^{- 1/2}}dx\) where \(t = \sqrt{1 + \frac{1}{x^{4}}}\)...

Evaluate ∫x−11(1+x4)−1/2dx\int_{}^{}{x^{- 11}(1 + x^{4})^{- 1/2}}dx where t=1+1x4t = \sqrt{1 + \frac{1}{x^{4}}}

A

12(t5+t3+t)+c\frac{1}{2}(t^{5} + t^{3} + t) + c

B

−12[t55−2t33+t]+c- \frac{1}{2}\left\lbrack \frac{t^{5}}{5} - \frac{2t^{3}}{3} + t \right\rbrack + c

C

12[t44+2t33+t]+c\frac{1}{2}\left\lbrack \frac{t^{4}}{4} + \frac{2t^{3}}{3} + t \right\rbrack + c

D

None of these

Answer

−12[t55−2t33+t]+c- \frac{1}{2}\left\lbrack \frac{t^{5}}{5} - \frac{2t^{3}}{3} + t \right\rbrack + c

Explanation

Solution

Here (m+1n+P)=[−11+14+12]=−3\left( \frac{m + 1}{n} + P \right) = \left\lbrack \frac{- 11 + 1}{4} + \frac{1}{2} \right\rbrack = - 3If we substitute then

1+1x4=t21 + \frac{1}{x^{4}} = t^{2} and −4x5dx=2tdt\frac{- 4}{x^{5}}dx = 2tdt

⇒I=∫dxx11(1+x4)1/2=∫dxx11.x2(1+1/x4)1/2I=∫dxx13(1+1/x4)1/2=14∫2tdtx8t=−12∫(t2−1)2dt=−12∫(t4−2t2+1)dt=−12[t55−2t33+t]+c,\Rightarrow I = \int_{}^{}\frac{dx}{x^{11}(1 + x^{4})^{1/2}} = \int_{}^{}\frac{dx}{x^{11}.x^{2}(1 + 1/x^{4})^{1/2}}I = \int_{}^{}{\frac{dx}{x^{13}(1 + 1/x^{4})^{1/2}} = \frac{1}{4}\int_{}^{}\frac{2tdt}{x^{8}t}} = - \frac{1}{2}\int_{}^{}{(t^{2} - 1)^{2}dt = \frac{- 1}{2}\int_{}^{}{(t^{4} - 2t^{2} + 1})dt} = \frac{- 1}{2}\left\lbrack \frac{t^{5}}{5} - \frac{2t^{3}}{3} + t \right\rbrack + c,

where t=1+1x4t = \sqrt{1 + \frac{1}{x^{4}}}