{x² d x}{≤ft(x²+2)≤ft(x²+5)}=

Correct Answer is : −23tan⁡−1(x2)+53tan⁡−1(x5)+c-{{2}}{3} ^{-1}≤ft({x}{{2}})+{{5}}{3} ^{-1}≤ft({x}{{5}})+c−32tan−1(2x)+35tan−1(5x)+c, where c is the constant of integration

-≤ft({{2}}{3} ^{-1} {x}{{2}}+{{5}}{3} ^{-1} {x}{{5}})+c, where c is the constant of integration

≤ft({{2}}{3} ^{-1} {x}{{2}}+{{5}}{3} ^{-1} {x}{{5}})+c, where c is the constant of integration

{{2}}{3} ^{-1}≤ft({x}{{2}})-{{5}}{3} ^{-1}≤ft({x}{{5}})+c, where c is the constant of integration

-{{2}}{3} ^{-1}≤ft({x}{{2}})+{{5}}{3} ^{-1}≤ft({x}{{5}})+c, where c is the constant of integration

{x² d x}{≤ft(x²+2)≤ft(x²+5)}=

Correct Answer is : −23tan⁡−1(x2)+53tan⁡−1(x5)+c-{{2}}{3} ^{-1}≤ft({x}{{2}})+{{5}}{3} ^{-1}≤ft({x}{{5}})+c−32tan−1(2x)+35tan−1(5x)+c, where c is the constant of integration

Test Index

MHTCET 15 Apr 2026 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int \frac{x^{2} d x}{\left(x^{2}+2\right)\left(x^{2}+5\right)}=

$-\left(\frac{\sqrt{2}}{3} \tan ^{-1} \frac{x}{\sqrt{2}}+\frac{\sqrt{5}}{3} \tan ^{-1} \frac{x}{\sqrt{5}}\right)+c$ , where c is the constant of integration
$\left(\frac{\sqrt{2}}{3} \tan ^{-1} \frac{x}{\sqrt{2}}+\frac{\sqrt{5}}{3} \tan ^{-1} \frac{x}{\sqrt{5}}\right)+c$ , where c is the constant of integration
$\frac{\sqrt{2}}{3} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{\sqrt{5}}{3} \tan ^{-1}\left(\frac{x}{\sqrt{5}}\right)+c$ , where c is the constant of integration
$-\frac{\sqrt{2}}{3} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{\sqrt{5}}{3} \tan ^{-1}\left(\frac{x}{\sqrt{5}}\right)+c$ , where c is the constant of integration

Validate

Solution:

Concept:

Use partial fractions to split the integrand into simpler denominators, then integrate using the standard arctangent formula.

Explanation:

Write the given fraction as:

\frac{x^{2}}{(x^{2}+2)(x^{2}+5)} = \frac{A}{x^{2}+2} + \frac{B}{x^{2}+5}

Multiplying by

(x^{2}+2)(x^{2}+5)

x^{2} = A(x^{2}+5) + B(x^{2}+2)

x^{2} = (A+B)x^{2} + (5A+2B)

Comparing coefficients:

A+B = 1

5A+2B = 0

From

A+B=1

, let

B = 1 - A

. Substitute into the second equation:

5A + 2(1-A) = 0

5A + 2 - 2A = 0

3A = -2 \Rightarrow A = -\frac{2}{3}

Then

B = 1 - \left(-\frac{2}{3}\right) = \frac{5}{3}

Thus the integrand becomes:

-\frac{2}{3}\cdot\frac{1}{x^{2}+2} + \frac{5}{3}\cdot\frac{1}{x^{2}+5}

Integrate term by term:

\int \frac{x^{2}\,dx}{(x^{2}+2)(x^{2}+5)} = -\frac{2}{3}\int \frac{dx}{x^{2}+2} + \frac{5}{3}\int \frac{dx}{x^{2}+5}

Use

\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)

= -\frac{2}{3}\cdot\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{5}{3}\cdot\frac{1}{\sqrt{5}}\tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c

Simplify:

= -\frac{\sqrt{2}}{3}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{\sqrt{5}}{3}\tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c

Answer:

Option D:

-\frac{\sqrt{2}}{3} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{\sqrt{5}}{3} \tan ^{-1}\left(\frac{x}{\sqrt{5}}\right)+c

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