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Find k from ∫(x³−8)/(x³√(x⁴−x³+2)) | JEE

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

Question
If ∫x3−8x3x4−x3+2 dx=kx4−x3+2x2+C\displaystyle\int \dfrac{x^{3}-8}{x^{3}\sqrt{x^{4}-x^{3}+2}}\,dx=\dfrac{k\sqrt{x^{4}-x^{3}+2}}{x^{2}}+C, then kk equals
A33
B22correct
C77
D44
Solution
Step 1: Factor x4x^{4} out of the radicand:
x4−x3+2=x4 ⁣(1−1x+2x4)  ⇒  x4−x3+2=x21−1x+2x4x^{4}-x^{3}+2=x^{4}\!\left(1-\dfrac{1}{x}+\dfrac{2}{x^{4}}\right)\;\Rightarrow\;\sqrt{x^{4}-x^{3}+2}=x^{2}\sqrt{1-\dfrac{1}{x}+\dfrac{2}{x^{4}}}
 Step 2: Divide the outer numerator by x3x^{3}: x3−8x3=1−8x3\dfrac{x^{3}-8}{x^{3}}=1-\dfrac{8}{x^{3}}. Combining with Step 1,
integrand=1−8x3x21−1x+2x4=1x2−8x51−1x+2x4\text{integrand}=\dfrac{1-\tfrac{8}{x^{3}}}{x^{2}\sqrt{1-\tfrac{1}{x}+\tfrac{2}{x^{4}}}}=\dfrac{\tfrac{1}{x^{2}}-\tfrac{8}{x^{5}}}{\sqrt{1-\tfrac{1}{x}+\tfrac{2}{x^{4}}}}
 Step 3: Substitute u=1−1x+2x4u=1-\dfrac{1}{x}+\dfrac{2}{x^{4}}. Then
du=(1x2−8x5)dxdu=\left(\dfrac{1}{x^{2}}-\dfrac{8}{x^{5}}\right)dx
 exactly the numerator. Step 4: Integrate:
I=∫duu=2u+CI=\int\dfrac{du}{\sqrt{u}}=2\sqrt{u}+C
 Step 5: Back-substitute u=x4−x3+2x4u=\dfrac{x^{4}-x^{3}+2}{x^{4}}, so u=x4−x3+2x2\sqrt{u}=\dfrac{\sqrt{x^{4}-x^{3}+2}}{x^{2}}:
I=2x4−x3+2x2+C  ⇒  k=2I=\dfrac{2\sqrt{x^{4}-x^{3}+2}}{x^{2}}+C\;\Rightarrow\; k=2
 Correct answer: (2)
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