Question

Prove that \[ \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) dx = 2x \tan^{-1}x - \log(1+x^2) + C. \]

Hint:

Whenever you see $\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)$, replace it with $2\tan^{-1}x$ using the double-angle identity for sine.

Answer 1

Step 1: Simplify the integrand using an identity.
We know that \[ \sin(2\tan^{-1}x) = \frac{2x}{1+x^2}. \] Thus, \[ \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x. \]

Step 2: Rewrite the integral.
\[ I = \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) dx = \int 2\tan^{-1}x \, dx. \]

Step 3: Use integration by parts.
Let \[ u = 2\tan^{-1}x, dv = dx. \] Then, \[ du = \frac{2}{1+x^2} dx, v = x. \] By parts formula: \[ I = uv - \int v \, du \] \[ I = 2x\tan^{-1}x - \int x \cdot \frac{2}{1+x^2} dx. \]

Step 4: Solve the remaining integral.
\[ \int \frac{2x}{1+x^2} dx = \ln(1+x^2). \] So, \[ I = 2x\tan^{-1}x - \ln(1+x^2) + C. \]

Final Answer: \[ \boxed{ \int \sin^{-1}\!\left(\frac{2x}{1+x^2}\right) dx = 2x\tan^{-1}x - \ln(1+x^2) + C } \]