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Primitive of (3x⁴−1)/(x⁴+x+1)² | JEE

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

Question

A primitive of

\dfrac{3x^{4}-1}{(x^{4}+x+1)^{2}}

with respect to

x

\dfrac{x}{x^{4}+x+1}+c

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

correct

\dfrac{x+1}{x^{4}+x+1}+c

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

Solution

Step 1: Divide numerator and denominator by

x^{2}

. Since

x^{4}+x+1=x^{2}\!\left(x^{2}+\dfrac{1}{x}+\dfrac{1}{x^{2}}\right)

, factoring instead about

x^{3}

gives the cleaner grouping

x^{4}+x+1=x\!\left(x^{3}+\dfrac{1}{x}+1\right)\;\Rightarrow\;(x^{4}+x+1)^{2}=x^{2}\!\left(x^{3}+\dfrac{1}{x}+1\right)^{2}

Hence

\dfrac{3x^{4}-1}{(x^{4}+x+1)^{2}}=\dfrac{3x^{2}-\tfrac{1}{x^{2}}}{\left(x^{3}+\tfrac{1}{x}+1\right)^{2}}

Step 2: Substitute

u=x^{3}+\dfrac{1}{x}+1

. Then

du=\left(3x^{2}-\dfrac{1}{x^{2}}\right)dx

exactly the numerator. Step 3: Integrate:

I=\int\dfrac{du}{u^{2}}=-\dfrac{1}{u}+c=-\dfrac{1}{x^{3}+\tfrac{1}{x}+1}+c

Step 4: Multiply numerator and denominator by

x

I=-\dfrac{x}{x^{4}+x+1}+c

(Verification:

\dfrac{d}{dx}\!\left(-\dfrac{x}{x^{4}+x+1}\right)=\dfrac{3x^{4}-1}{(x^{4}+x+1)^{2}}

.) Correct answer: (2)

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