Question

If ∫ \(\frac{x^{49} Tan^{-1} (x^{50})}{(1+x^{100})}\)dx = k(Tan^-1 (x⁵⁰))² + c, then k =

• A. \(\frac{-1}{100}\)
• B. \(\frac{1}{50}\)
• C. \(\frac{-1}{50}\)
• D. \(\frac{1}{100}\)

Answer 1

The Correct Option is D

To solve the problem, we need to find the constant $k$ in the integral $\int \frac{x^{49} \arctan(x^{50})}{1 + x^{100}} dx = k (\arctan(x^{50}))^2 + C$.

1. Set Up the Substitution:
Let $u = \arctan(x^{50})$.
Then $du = \frac{1}{1 + (x^{50})^2} \cdot 50 x^{49} dx = \frac{50 x^{49}}{1 + x^{100}} dx$.
Thus, $\frac{du}{50} = \frac{x^{49}}{1 + x^{100}} dx$.

2. Rewrite the Integral:
Substitute into the integral:
$I = \int \frac{x^{49} \arctan(x^{50})}{1 + x^{100}} dx = \int u \cdot \frac{du}{50} = \frac{1}{50} \int u du$.

3. Evaluate the Integral:
Integrate:
$\frac{1}{50} \int u du = \frac{1}{50} \cdot \frac{u^2}{2} + C = \frac{1}{100} u^2 + C$.
Substitute back $u = \arctan(x^{50})$:
$\frac{1}{100} (\arctan(x^{50}))^2 + C$.

4. Determine $k$:
The integral is given as $k (\arctan(x^{50}))^2 + C$.
Comparing with $\frac{1}{100} (\arctan(x^{50}))^2 + C$, we find $k = \frac{1}{100}$.

Final Answer:
The value of $k$ is $\frac{1}{100}$.