Q88 / 150 · MHT CET 2025 (25 Apr Shift 2) · Mathematics
$\int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} \mathrm{~d} x$ equals
- Correct answerA$\frac{\sin \left(\tan ^{-1} x^5\right)}{5}+c \quad, \quad$ where $c$ is the constant of integration
- B$x^4 \sin \left(\tan ^{-1} x^5\right)+\mathrm{c}, \quad$ where c is the constant of integration
- C$\frac{\sin \left(\tan ^{-1} x^5\right)}{4}+c \quad, \quad$ where $c$ is the constant of integration
- D$\cos \left(\tan ^{-1} x^5\right)+\mathrm{c}, \quad$ where c is the constant of integration
Solution
Evaluate $I = \int \frac{x^4 \cos (\tan^{-1} x^5)}{1+x^{10}} d x$ using substitution.
Let $u = \tan^{-1} x^5$, so $\tan u = x^5$.
Differentiate: $\frac{du}{dx} = \frac{1}{1+(x^5)^2} \cdot 5x^4 = \frac{5x^4}{1+x^{10}}$.
Rewriting gives $\frac{x^4}{1+x^{10}} dx = \frac{1}{5} du$.
Substitute into $I$:
$I = \int \cos u \cdot \frac{1}{5} du = \frac{1}{5} \sin u + C$.
Back-substitute $u = \tan^{-1} x^5$ to obtain $I = \frac{1}{5} \sin(\tan^{-1} x^5) + C$.
The result corresponds with option A.
Hence, the correct answer is option: 'A'.