{x³}{x^{4}+5 x²+4} d x=

Correct Answer is : 13log⁡((x2+4)2x2+1)+c1/3 ≤ft({≤ft(x²+4)²}{{x²+1}})+c31log(x2+1(x2+4)2)+c, where c is the constant of integration

1/3 ≤ft({≤ft(x²+4)²}{{x²+1}})+c, where c is the constant of integration

≤ft({≤ft(x²+4)²}{{x²+1}})+c, where c is the constant of integration

3 ≤ft({≤ft(x²+4)²}{{x²+1}})+c, where c is the constant of integration

2/3 ≤ft({≤ft(x²+4)²}{{x²+1}})+c, where c is the constant of integration

Correct Answer is : 13log⁡((x2+4)2x2+1)+c1/3 ≤ft({≤ft(x²+4)²}{{x²+1}})+c31log(x2+1(x2+4)2)+c, where c is the constant of integration

Test Index

MHTCET 22 Apr 2025 Shift 2 Paper

Question : 114 of 150

Marks: +1, -0

\int \frac{x^{3}}{x^{4}+5 x^{2}+4} d x=

$\frac{1}{3} \log \left(\frac{\left(x^{2}+4\right)^{2}}{\sqrt{x^{2}+1}}\right)+c$ , where c is the constant of integration
$\log \left(\frac{\left(x^{2}+4\right)^{2}}{\sqrt{x^{2}+1}}\right)+c$ , where c is the constant of integration
$3 \log \left(\frac{\left(x^{2}+4\right)^{2}}{\sqrt{x^{2}+1}}\right)+c$ , where c is the constant of integration
$\frac{2}{3} \log \left(\frac{\left(x^{2}+4\right)^{2}}{\sqrt{x^{2}+1}}\right)+c$ , where c is the constant of integration

Solution:

Concept:

Use the substitution

x^2 = t

and then apply partial fractions to integrate.

Explanation:

Let

I = \int \frac{x^3}{x^4+5x^2+4}\,dx

Put

x^2 = t

, so

2x\,dx = dt

Then

x^3\,dx = x^2 \cdot x\,dx = \frac{t}{2}\,dt

Hence,

I = \frac{1}{2}\int \frac{t}{t^2+5t+4}\,dt

Factor the denominator:

t^2+5t+4 = (t+1)(t+4)

Write

\frac{t}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}

t = A(t+4) + B(t+1)

Putting

t = -1

gives

A = -\frac{1}{3}

Putting

t = -4

gives

B = \frac{4}{3}

Therefore,

I = \frac{1}{2}\left[-\frac{1}{3}\int \frac{dt}{t+1} + \frac{4}{3}\int \frac{dt}{t+4}\right]

This simplifies to

I = -\frac{1}{6}\log(t+1) + \frac{2}{3}\log(t+4) + c

Rewrite as

I = \frac{1}{3}\left[-\frac{1}{2}\log(t+1) + 2\log(t+4)\right] + c

Using logarithm rules:

-\frac{1}{2}\log(t+1) = -\log\sqrt{t+1}

and

2\log(t+4) = \log(t+4)^2

I = \frac{1}{3}\log\left(\frac{(t+4)^2}{\sqrt{t+1}}\right) + c

Substitute

t = x^2

to get

I = \frac{1}{3}\log\left(\frac{(x^2+4)^2}{\sqrt{x^2+1}}\right) + c

Answer:

\frac{1}{3} \log \left(\frac{\left(x^{2}+4\right)^{2}}{\sqrt{x^{2}+1}}\right)+c

, where

c

is the constant of integration.

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