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Integral of x⁻⁷(1+x⁴)^(−1/2) | JEE

JEE Maths question with a full step-by-step solution.

Question
∫x−7(1+x4)−1/2 dx\displaystyle\int x^{-7}(1+x^{4})^{-1/2}\,dx is equal to
Ax4+1 (2x4−1)6x6+C\dfrac{\sqrt{x^{4}+1}\,(2x^{4}-1)}{6x^{6}}+Ccorrect
Bx4+1 (2x4+1)6x6+C\dfrac{\sqrt{x^{4}+1}\,(2x^{4}+1)}{6x^{6}}+C
Cx4+1 (2x4+1)4x6+C\dfrac{\sqrt{x^{4}+1}\,(2x^{4}+1)}{4x^{6}}+C
Dx4+1 (2x4−1)4x6+C\dfrac{\sqrt{x^{4}+1}\,(2x^{4}-1)}{4x^{6}}+C
Solution
Step 1: Pull x4x^{4} out of the bracket. Since 1+x4=x4(x−4+1)1+x^{4}=x^{4}(x^{-4}+1),
(1+x4)−1/2=x−2(x−4+1)−1/2  ⇒  x−7(1+x4)−1/2=x−9(x−4+1)−1/2(1+x^{4})^{-1/2}=x^{-2}(x^{-4}+1)^{-1/2}\;\Rightarrow\; x^{-7}(1+x^{4})^{-1/2}=x^{-9}(x^{-4}+1)^{-1/2}
⇒I=∫x−9(x−4+1)−1/2 dx\Rightarrow I=\int x^{-9}(x^{-4}+1)^{-1/2}\,dx
 Step 2: Substitute t2=x−4+1t^{2}=x^{-4}+1. Differentiating, 2t dt=−4x−5 dx⇒x−5 dx=−t2 dt2t\,dt=-4x^{-5}\,dx\Rightarrow x^{-5}\,dx=-\dfrac{t}{2}\,dt. Also x−4=t2−1x^{-4}=t^{2}-1, so
x−9 dx=x−4⋅x−5 dx=(t2−1) ⁣(−t2)dt,(x−4+1)−1/2=(t2)−1/2=1tx^{-9}\,dx=x^{-4}\cdot x^{-5}\,dx=(t^{2}-1)\!\left(-\dfrac{t}{2}\right)dt,\qquad (x^{-4}+1)^{-1/2}=(t^{2})^{-1/2}=\dfrac{1}{t}
 Step 3: Assemble the integral:
I=∫1t (t2−1) ⁣(−t2)dt=−12∫(t2−1) dtI=\int\dfrac{1}{t}\,(t^{2}-1)\!\left(-\dfrac{t}{2}\right)dt=-\dfrac{1}{2}\int(t^{2}-1)\,dt
 Step 4: Integrate:
=−12(t33−t)+C=t2−t36+C=-\dfrac{1}{2}\left(\dfrac{t^{3}}{3}-t\right)+C=\dfrac{t}{2}-\dfrac{t^{3}}{6}+C
 Step 5: Back-substitute t=x−4+1=x4+1x2t=\sqrt{x^{-4}+1}=\dfrac{\sqrt{x^{4}+1}}{x^{2}}:
I=x4+12x2−(x4+1)3/26x6+C=x4+1 [3x4−(x4+1)]6x6+C=x4+1 (2x4−1)6x6+CI=\dfrac{\sqrt{x^{4}+1}}{2x^{2}}-\dfrac{(x^{4}+1)^{3/2}}{6x^{6}}+C=\dfrac{\sqrt{x^{4}+1}\,\big[3x^{4}-(x^{4}+1)\big]}{6x^{6}}+C=\dfrac{\sqrt{x^{4}+1}\,(2x^{4}-1)}{6x^{6}}+C
 Correct answer: (1)
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