If I=∫tan−1⁡(1+x)dx, then I equals

I=∫tan−1⁡(1+x)dx put 1+x=tan⁡θ ⇒12xdx=sec2θdθ ⇒dx=2tan⁡θ-1sec2θdθ I=∫θ.2tan⁡θ-1sec2θdθ =θ.tan⁡θ-12-∫1.tan⁡θ-12dθ =θ.tan⁡θ-12-∫sec2θ-2tanθdθ =θ.tan⁡θ-12-tanθ-2logsecθ+c =x2tan−1⁡(1+x)−x+log⁡(x+2x+2)+C

If I=∫tan−1⁡(1+x)dx, then I equals

I=∫tan−1⁡(1+x)dx put 1+x=tan⁡θ ⇒12xdx=sec2θdθ ⇒dx=2tan⁡θ-1sec2θdθ I=∫θ.2tan⁡θ-1sec2θdθ =θ.tan⁡θ-12-∫1.tan⁡θ-12dθ =θ.tan⁡θ-12-∫sec2θ-2tanθdθ =θ.tan⁡θ-12-tanθ-2logsecθ+c =x2tan−1⁡(1+x)−x+log⁡(x+2x+2)+C

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If $I = \int \tan^{- 1} (1 + \sqrt{x}) dx$ , then I equals

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none of these

$x^{2} \tan^{- 1} (1 + \sqrt{x}) - \sqrt{x} + \log (x + 2 \sqrt{x} + 2) + C$

$(\sqrt{x} - 1)^{2} \tan^{- 1} (1 + \sqrt{x}) - \sqrt{x} + \log (\sqrt{x} + 1) + C$

$(2 \sqrt{2} + 1) \tan^{- 1} (1 + \sqrt{x}) + \sqrt{x} - \log (x + 2 \sqrt{x} + 2) + C$

answer is A.

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Detailed Solution

$I = \int \tan^{- 1} (1 + \sqrt{x}) dx p u t 1 + \sqrt{x} = \tan θ \Rightarrow \frac{1}{2 \sqrt{x}} dx = \sec^{2} θdθ \Rightarrow dx = 2 (\tan θ - 1) \sec^{2} θdθ I = \int θ . 2 (\tan θ - 1) \sec^{2} θdθ = θ . {(\tan θ - 1)}^{2} - \int 1 . {(\tan θ - 1)}^{2} dθ = θ . {(\tan θ - 1)}^{2} - \int [\sec^{2} θ - 2 tanθ] dθ = θ . {(\tan θ - 1)}^{2} - (tanθ - 2 \log |secθ|) + c = x^{2} \tan^{- 1} (1 + \sqrt{x}) - \sqrt{x} + \log (x + 2 \sqrt{x} + 2) + C$