Question

$\int_{}^{}\frac{\mathbf{dx}}{\mathbf{(x}\mathbf{-}\mathbf{3)}\sqrt{\mathbf{x + 1}}}\mathbf{=}$

• A. $\frac{1}{2}\log\left( \frac{\sqrt{x + 1} + 2}{\sqrt{x + 1} - 2} \right) + c$
• B. $\frac{1}{2}\log\left( \frac{\sqrt{x + 1} + 2}{\sqrt{x + 1} - 2} \right) + c$
• C. $\frac{1}{2}\log\left( \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} \right) + c$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}\log\left( \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} \right) + c$

Answer 2

Put $x + 1 = t^{2}$ $\Rightarrow dx = 2tdt$

$\int_{}^{}\frac{dx}{(x - 3)\sqrt{x + 1}} = \int_{}^{}\frac{2tdt}{(t^{2} - 4)t}$ $\left( \because x + 1 = t^{2} \Rightarrow x = t^{2} - 1 \Rightarrow x - 3 = t^{2} - 4 \right)$

$= 2\int_{}^{}\frac{dt}{t^{2} - 2^{2}}$ $= 2.\frac{1}{2.2}\log\left( \frac{t - 2}{t + 2} \right) + c = \frac{1}{2}\log\left( \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} \right) + c$.