Q85 / 150 · MHT CET 2025 (23 Apr Shift 2) · Mathematics

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

  • Correct answer
    A
    $\frac{1}{2} \tan ^{-1} x^2+c$, where $c$ is the constant of integration
  • B
    $2 \tan ^{-1} x+c$, where $c$ is the constant of integration
  • C
    $\frac{1}{2} \tan ^{-1} x+c, \quad$ where $c$ is the constant of integration
  • D
    $\tan ^{-1} x^2+c, \quad$ where $c$ is the constant of integration

Solution

The integral $I = \int \frac{x}{1+x^4} dx$ can be evaluated using substitution.

Let $u = x^2$, so $du = 2x dx$ and $x dx = \frac{1}{2} du$.

Substituting yields $I = \frac{1}{2} \int \frac{1}{1+u^2} du$.

Recognizing the standard anti-derivative gives $I = \frac{1}{2} \tan^{-1} u + c$.

Substituting back $u = x^2$ produces the final result:

$I = \frac{1}{2} \tan^{-1} x^2 + c$

Hence, the correct answer is option: 'A'.

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