Put, x = t^2 dx = 2t dt ∫√(x)x + 1 dx = ∫tt^2 + 1 2t dt = 2∫t^2t^2 + 1 dt = 2∫t^2 + 1 1t^2 + 1 dt = 2(∫ 1 dt ∫1t^2 + 1)dt = 2 ...

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Derivative of Standard Functions

Evaluate ∫√...

Question

Evaluate $\int \frac{\sqrt{x}}{x + 1} d x$

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Solution

Put, $x = t^{2} ⟹ d x = 2 t d t$

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

$= 2 \int \frac{t^{2}}{t^{2} + 1} d t$

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

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

$= 2 (t - {tan}^{- 1} t) + c$

$= 2 \sqrt{x} - 2 {tan}^{- 1} \sqrt{x} + c$ (Ans)

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