Mathematics

Mathematics

Problem

∫tan⁡−12x1−x2 dx=\int \tan^{-1}\frac{2x}{1-x^2}\,dx = (a) xtan⁡−1x+cx\tan^{-1}x + c (b) xtan⁡−1x−log⁡(1+x2)+cx\tan^{-1}x - \log(1+x^2) + c (c) 2xtan⁡−1x+log⁡(1+x2)+c2x\tan^{-1}x + \log(1+x^2) + c (d) 2xtan⁡−1x−log⁡(1+x2)+c2x\tan^{-1}x - \log(1+x^2) + c (a) xtan⁡−1x+cx\tan^{-1}x + c (b) xtan⁡−1x−log⁡(1+x2)+cx\tan^{-1}x - \log(1+x^2) + c (c) 2xtan⁡−1x+log⁡(1+x2)+c2x\tan^{-1}x + \log(1+x^2) + c (d) 2xtan⁡−1x−log⁡(1+x2)+c2x\tan^{-1}x - \log(1+x^2) + c
Intuitive explanation

The key is recognizing that the argument 2x1−x2\frac{2x}{1-x^2} matches the tangent double-angle formula: tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}. If you set tan⁡θ=x\tan\theta = x, then θ=tan⁡−1x\theta = \tan^{-1}x, and the integrand simplifies to tan⁡−1(tan⁡(2tan⁡−1x))=2tan⁡−1x\tan^{-1}(\tan(2\tan^{-1}x)) = 2\tan^{-1}x. Now you have a much simpler integral to evaluate using integration by parts.

Intuitive explanation

The key is recognizing that the argument 2x1−x2\frac{2x}{1-x^2} matches the tangent double-angle formula: tan⁡(2θ)=2tan⁡θ1−tan⁡2θ\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}. If you set tan⁡θ=x\tan\theta = x, then θ=tan⁡−1x\theta = \tan^{-1}x, and the integrand simplifies to tan⁡−1(tan⁡(2tan⁡−1x))=2tan⁡−1x\tan^{-1}(\tan(2\tan^{-1}x)) = 2\tan^{-1}x. Now you have a much simpler integral to evaluate using integration by parts.

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