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Question

tan-1x dx is equal to(a) (x+1)tan-1x-x+C (b) x tan-1x-x+C(c) x-x tan-1 x+C (d) x-(x+1)tan-1x+C

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Solution


I=tan-1xdx

Put x=tan2θ

dx=2tanθsec2θdθ

I=tan-1tan2θ×2tanθsec2θdθ

I=2θ×tanθsec2θdθ

I=2θtanθsec2θdθ-ddθθtanθsec2θdθdθ

I=2θ×tan2θ2-tan2θ2dθ

I=θtan2θ-sec2θ-1dθ

I=θtan2θ-sec2θdθ+dθ

I=θtan2θ-tanθ+θ+C

I=tan-1x×x-x+tan-1x+C x=tan2θtanθ=xθ=tan-1x

I=x+1tan-1x-x+C

Hence, the correct answer is option (a).

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