Question

Evaluate \( \int \tan^{-1} x \, dx \).

Hint:

Use integration by parts for \( \tan^{-1} x \); simplify the resulting integral with substitution.

Answer 1

Use integration by parts: Let \( u = \tan^{-1} x \), \( dv = dx \). Then \( du = \frac{1}{1 + x^2} dx \), \( v = x \). \[ \int \tan^{-1} x \, dx = x \tan^{-1} x - \int \frac{x}{1 + x^2} \, dx. \] For \( \int \frac{x}{1 + x^2} \, dx \), let \( t = 1 + x^2 \), \( dt = 2x \, dx \), \( dx = \frac{dt}{2x} \), so: \[ \int \frac{x}{1 + x^2} \, dx = \int \frac{x}{t} \cdot \frac{dt}{2x} = \frac{1}{2} \int \frac{1}{t} \, dt = \frac{1}{2} \ln |t| = \frac{1}{2} \ln (1 + x^2). \] \[ \int \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \ln (1 + x^2) + c. \] Answer: \( x \tan^{-1} x - \frac{1}{2} \ln (1 + x^2) + c \).