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Question

Mathematics Question on integral

∫tan−1x dx\int tan^{-1} \sqrt{x}\,dx is equal to

A

(x+1)tan−1x−x+C\left(x + 1\right)tan^{-1}\sqrt{x} -\sqrt{x} + C

B

x tan−1x−x+Cx\,tan^{-1}\sqrt{x} -\sqrt{x} + C

C

x−x tan−1x+C\sqrt{x} - x\, tan^{-1}\sqrt{x} + C

D

x−(x+1)tan−1x+C\sqrt{x} - (x+1) tan^{-1}\sqrt{x} + C

Answer

(x+1)tan−1x−x+C\left(x + 1\right)tan^{-1}\sqrt{x} -\sqrt{x} + C

Explanation

Solution

We have, I=∫1⋅tan−1x dxI = \int 1\cdot tan^{-1}\sqrt{x}\,dx ⇒I=tan−1x⋅(x)−∫11+x×12x×x dx\Rightarrow I= tan^{-1}\sqrt{x} \cdot\left(x\right) -\int\frac{1}{ 1+x} \times \frac{1}{2\sqrt{x}} \times x \, dx =x tan−1x−∫x(1+x) 2 xdx = x\,tan^{-1}\sqrt{x}-\int\frac{x}{ \left(1+x\right) \,2\,\sqrt{x}} dx =x tan−1x−∫(1+x(1+x) 2 x−1(1+x) 2 x)dx = x\, tan^{-1} \sqrt{x} -\int\left(\frac{1+x}{\left(1+x\right)\, 2\,\sqrt{x}} - \frac{1}{\left(1+x\right)\,2\,\sqrt{x}}\right) dx =x tan−1x−∫dx2 x+∫dx2 x(1+x)= x\, tan^{-1} \sqrt{x}- \int\frac{dx}{2\,\sqrt{x}} + \int \frac{dx}{2\,\sqrt{x} \left(1+x\right)} =x tan−1x−x+tan−1x+C= x\,tan^{-1}\sqrt{x} - \sqrt{x} + tan^{-1}\sqrt{x} +C =(x+1)tan−1x−x+C =\left(x+1\right)tan^{-1} \sqrt{x} - \sqrt{x} + C