Indefinite IntegrationmediumFree

Integral of (x³−1)/√(x⁶+2x³) | JEE

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

Question

\displaystyle\int \dfrac{(x^{3}-1)\,dx}{\sqrt{x^{6}+2x^{3}}}

equals

\sqrt{x^{4}+2x}+C

\sqrt{\dfrac{x^{3}+2}{x^{2}}}+C

\sqrt{\dfrac{x^{6}+2x^{3}}{x^{3}}}+C

\sqrt{\dfrac{x^{3}+2}{x}}+C

correct

Solution

Step 1: Factor

x^{4}

from the radicand to expose a usable inner function:

x^{6}+2x^{3}=x^{4}\!\left(x^{2}+\dfrac{2}{x}\right)\;\Rightarrow\;\sqrt{x^{6}+2x^{3}}=x^{2}\sqrt{x^{2}+\dfrac{2}{x}}

Step 2: Divide the numerator by

x^{2}

\dfrac{x^{3}-1}{x^{2}}=x-\dfrac{1}{x^{2}}\;\Rightarrow\; I=\int\dfrac{x-\tfrac{1}{x^{2}}}{\sqrt{x^{2}+\tfrac{2}{x}}}\,dx

Step 3: Substitute

t=x^{2}+\dfrac{2}{x}

. Then

dt=\left(2x-\dfrac{2}{x^{2}}\right)dx=2\left(x-\dfrac{1}{x^{2}}\right)dx\;\Rightarrow\;\left(x-\dfrac{1}{x^{2}}\right)dx=\dfrac{dt}{2}

Step 4: The integral becomes

I=\int\dfrac{dt/2}{\sqrt{t}}=\dfrac{1}{2}\cdot 2\sqrt{t}=\sqrt{t}+C=\sqrt{x^{2}+\dfrac{2}{x}}+C

Step 5: Write under a single radical:

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

Correct answer: (4)

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