Correct Answer is : 12tan⁡−1x2+c1/2 ^{-1} x²+c21tan−1x2+c, where ccc is the constant of integration

1/2 ^{-1} x²+c, where c is the constant of integration

2 ^{-1} x+c, where c is the constant of integration

1/2 ^{-1} x+c, where c is the constant of integration

^{-1} x²+c^{} where c is the constant of integration

Correct Answer is : 12tan⁡−1x2+c1/2 ^{-1} x²+c21tan−1x2+c, where ccc is the constant of integration

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MHTCET 23 Apr 2025 Shift 2 Paper

Question : 135 of 150

Marks: +1, -0

\int \frac{x}{1+x^{4}} d x=

Solution:

Concept:

Use the substitution

x^2 = t

to reduce the integral to a standard form.

Explanation:

Let

\mathrm{I} = \int \frac{x}{1+x^4} \, dx

Put

x^2 = t

, so

2x \, dx = dt

Hence

x \, dx = \frac{dt}{2}

Substituting into the integral:

\mathrm{I} = \frac{1}{2} \int \frac{dt}{1+t^2}

We know

\int \frac{dt}{1+t^2} = \tan^{-1}(t) + c

Therefore,

\mathrm{I} = \frac{1}{2} \tan^{-1}(t) + c

Now replace

t = x^2

\mathrm{I} = \frac{1}{2} \tan^{-1}(x^2) + c

Answer:

\frac{1}{2} \tan^{-1}(x^2) + c

, which matches option A.

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