Question

a. \\(\int \frac{tan^{-1}(x)}{1 + x^2} dx\\)

Explanation:

Step1: Substitute \( u = \tan^{-1}(x) \)

Then, find \( du \). The derivative of \( \tan^{-1}(x) \) with respect to \( x \) is \( \frac{1}{1 + x^2} \), so \( du = \frac{1}{1 + x^2}dx \).

Step2: Rewrite the integral

The integral becomes \( \int u \, du \) (since \( \frac{\tan^{-1}(x)}{1 + x^2}dx = u \, du \)).

Step3: Integrate \( \int u \, du \)

Using the power rule for integration \( \int x^n dx = \frac{x^{n + 1}}{n + 1} + C \) (here \( n = 1 \)), we get \( \frac{u^2}{2} + C \).

Step4: Substitute back \( u = \tan^{-1}(x) \)

The result is \( \frac{(\tan^{-1}(x))^2}{2} + C \).

Answer:

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