Question

Find : \( \int \tan^{-1}\left(\frac{1 - x}{1 + x}\right) \, dx \)

Hint:

Alternatively, substituting \( x = \tan\theta \) directly simplifies the inner fraction to \( \tan(\frac{\pi}{4} - \theta) \), transforming the entire inverse function expression into a simple linear expression \( \frac{\pi}{4} - \theta \) right from the start.

Answer 1

Concept: Before applying integration techniques like Integration by Parts, we can simplify the inverse trigonometric expression using standard trigonometric identity substitutions or inverse tangent properties: \[ \tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B \]

Step 1: Simplify the inverse trigonometric function expression.
Let the integrand factor be split using the property formula, where \( A = 1 \) and \( B = x \): \[ \tan^{-1}\left(\frac{1 - x}{1 + 1 \cdot x}\right) = \tan^{-1}(1) - \tan^{-1}(x) \] Since \( \tan^{-1}(1) = \frac{\pi}{4} \), the integral can be rewritten as: \[ I = \int \left( \frac{\pi}{4} - \tan^{-1}x \right) dx = \frac{\pi}{4}\int dx - \int \tan^{-1}x \, dx \] \[ I = \frac{\pi x}{4} - \int \tan^{-1}x \, dx \]

Step 2: Solve the remaining integral using Integration by Parts.
To compute \( I_1 = \int \tan^{-1}x \, dx \), we treat the integrand as a product with 1: \( \int (\tan^{-1}x \cdot 1) \, dx \). Using the ILATE rule, choose:
• First function \( u = \tan^{-1}x \implies du = \frac{1}{1 + x^2} \, dx \)
• Second function \( v = 1 \implies \int v \, dx = x \) Applying the Integration by Parts formula \( \int u \, dv = uv - \int v \, du \): \[ I_1 = x\tan^{-1}x - \int \frac{x}{1 + x^2} \, dx \]

Step 3: Integrate the algebraic fraction component.
For the remaining integral \( \int \frac{x}{1 + x^2} \, dx \), multiply and divide by 2 so the numerator perfectly reflects the denominator derivative: \[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2}\int \frac{2x}{1 + x^2} \, dx = \frac{1}{2}\log|1 + x^2| \] Thus, the value of \( I_1 \) is: \[ I_1 = x\tan^{-1}x - \frac{1}{2}\log|1 + x^2| \]

Step 4: Combine parts to write the final integral solution.
Substitute \( I_1 \) back into our primary expression setup: \[ I = \frac{\pi x}{4} - \left( x\tan^{-1}x - \frac{1}{2}\log|1 + x^2| \right) + C \] \[ I = \frac{\pi x}{4} - x\tan^{-1}x + \frac{1}{2}\log|1 + x^2| + C \]