tan - 1 x ? d x is equal to:

Question

$\int \tan^{- 1} \sqrt{x} d x$ is equal to:

Smart Achievers · Accepted Answer

(1)

Ans. $(x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C$

Explanation:

Let $I = \int 1 \cdot \tan^{- 1} \sqrt{x} d x$

$= \tan^{- 1} \sqrt{x} \cdot x - \int \frac{1}{1 + {(\sqrt{x})}^{2}} \cdot \frac{1}{2 \sqrt{x}} \cdot x d x$

$= x \tan^{- 1} \sqrt{x} - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} d x$

Put $x = t^{2} \Rightarrow d x = 2 t d t$

$∴$ $I = x \tan^{- 1} \sqrt{x} - \frac{1}{2} \int \frac{t \cdot 2 t}{1 + t^{2}} d t$

$= x \tan^{- 1} \sqrt{x} - \int \frac{t^{2}}{1 + t^{2}} d t$

$= x \tan^{- 1} \sqrt{x} - \int (1 - \frac{1}{1 + t^{2}}) d t$

$= x \tan^{- 1} \sqrt{x} - t + \tan^{- 1} t + C$

$= x \tan^{- 1} \sqrt{x} - \sqrt{x} + \tan^{- 1} \sqrt{x} + C$ $(\begin{matrix} ∵ x = t^{2} \\ ∴ t = \sqrt{x} \end{matrix})$

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

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