{x^{4}≤ft(x¹0}-1)}{x²0}+3 x¹0}+1} d x=·s

Correct Answer is : 15tan⁡−1(x5+1x5)+c1/5 ^{-1}≤ft(x^{5}+{1}{x^{5}})+c51tan−1(x5+x51)+c

1/5 ^{-1}≤ft(x^{5}+{1}{x^{5}})+c

1/10 ^{-1}≤ft(x¹0}+{1}{x¹0}})+c

Correct Answer is : 15tan⁡−1(x5+1x5)+c1/5 ^{-1}≤ft(x^{5}+{1}{x^{5}})+c51tan−1(x5+x51)+c

Test Index

MHTCET 20 Apr 2026 Shift 1 Paper

Question : 128 of 150

Marks: +1, -0

\int \frac{x^{4}\left(x^{10}-1\right)}{x^{20}+3 x^{10}+1} d x=\cdots

Solution:

Concept:

Use the derivative of

\tan^{-1}\left(x^5+\frac{1}{x^5}\right)

Explanation:

Let

u=x^5+\frac{1}{x^5}

Then

\frac{du}{dx}=5x^4-\frac{5}{x^6}=\frac{5(x^{10}-1)}{x^6}

Also,

1+u^2=1+\left(x^5+\frac{1}{x^5}\right)^2=x^{10}+3+\frac{1}{x^{10}}=\frac{x^{20}+3x^{10}+1}{x^{10}}

\frac{d}{dx}\tan^{-1}u=\frac{du/dx}{1+u^2}

Substituting the values:

\frac{5(x^{10}-1)}{x^6}\cdot\frac{x^{10}}{x^{20}+3x^{10}+1}=\frac{5x^4(x^{10}-1)}{x^{20}+3x^{10}+1}

Hence the given integrand is

\frac{1}{5}

times this derivative.

Therefore,

\int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1}dx=\frac{1}{5}\tan^{-1}\left(x^5+\frac{1}{x^5}\right)+c

Answer:

Option B:

\frac{1}{5}\tan^{-1}\left(x^5+\frac{1}{x^5}\right)+c

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