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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

\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

k

equals

3

2

correct

7

4

Solution

Step 1: Factor

x^{4}

out of the radicand:

x^{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

x^{3}

\dfrac{x^{3}-8}{x^{3}}=1-\dfrac{8}{x^{3}}

. Combining with Step 1,

\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-\dfrac{1}{x}+\dfrac{2}{x^{4}}

. Then

du=\left(\dfrac{1}{x^{2}}-\dfrac{8}{x^{5}}\right)dx

exactly the numerator. Step 4: Integrate:

I=\int\dfrac{du}{\sqrt{u}}=2\sqrt{u}+C

Step 5: Back-substitute

u=\dfrac{x^{4}-x^{3}+2}{x^{4}}

, so

\sqrt{u}=\dfrac{\sqrt{x^{4}-x^{3}+2}}{x^{2}}

I=\dfrac{2\sqrt{x^{4}-x^{3}+2}}{x^{2}}+C\;\Rightarrow\; k=2

Correct answer: (2)

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