Value of \(\int\displaystyle\frac{2}{(x-3)\sqrt{x+1}}\,dx\) is: (Here C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\log\left|\displaystyle\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+C\)
Let \(t = \sqrt{x+1}\) (substitution for the root). Then
\(x = t^2 - 1,\quad dx = 2t\,dt,\quad x - 3 = t^2 - 4\)
\( \displaystyle \int \frac{2}{(x-3)\sqrt{x+1}}\,dx = \int \frac{2 \cdot 2t}{(t^2 - 4)\,t}\,dt \)
\( \displaystyle = 4\int \frac{dt}{t^2 - 2^2} \)
By the standard integral \(\int \frac{dt}{t^2 - a^2} = \frac{1}{2a}\log\left|\frac{t-a}{t+a}\right| + C\), with a = 2:
\( \displaystyle = 4 \times \frac{1}{4}\log\left|\frac{t-2}{t+2}\right| + C \)
\( \displaystyle = \log\left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right| + C \)