{d x}{x≤ft(x³+1)}=

Correct Answer is : log⁡(x3x3+13)+c ≤ft([3]{{x³}{x³+1}})+clog(3x3+1x3)+c, where c is the constant of integration

≤ft({x³}{x³+1})+c, where c is the constant of integration

1/3 ≤ft([3]{{x³}{x³+1}})+c, where c is the constant of integration

≤ft([3]{{x³}{x³+1}})+c, where c is the constant of integration

1/3 ≤ft({x³+1}{x³})+c, where c is the constant of integration

Correct Answer is : log⁡(x3x3+13)+c ≤ft([3]{{x³}{x³+1}})+clog(3x3+1x3)+c, where c is the constant of integration

Test Index

MHTCET 25 Apr 2025 Shift 2 Paper

Question : 134 of 150

Marks: +1, -0

\int \frac{d x}{x\left(x^{3}+1\right)}=

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

Solution:

Concept:

Use substitution to reduce the integrand to a partial fraction form.

Explanation:

Multiply numerator and denominator by

x^2

so that the derivative of

x^3

appears.

Let

I=\int \frac{dx}{x(x^3+1)} = \int \frac{x^2}{x^3(x^3+1)}\,dx

Put

x^3=t

, then

3x^2\,dx=dt

I=\frac{1}{3}\int \frac{dt}{t(t+1)}

Using partial fractions:

\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}

Thus

I=\frac{1}{3}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt = \frac{1}{3}\left(\ln|t|-\ln|t+1|\right)+c

Simplify:

I=\frac{1}{3}\ln\left|\frac{t}{t+1}\right|+c

Substitute

t=x^3

I=\frac{1}{3}\ln\left|\frac{x^3}{x^3+1}\right|+c

This equals

\ln\left(\sqrt[3]{\frac{x^3}{x^3+1}}\right)+c

Answer:

Option C:

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

, where

c

is the constant of integration.

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