$\frac{-1}{2[1(+\sqrt{x})^6]}\left(\frac{6\sqrt{x}}{1+\sqrt{x}}+1\right) + c$
$\frac{-1}{2[1(+\sqrt{x})^7]}(7\sqrt{x} + 1) + c$
Either (A) and (B)
None of these
Step-by-Step Solution
Key Concept: Substitution $u = 1 + \sqrt{x}$ transforms the integral into a rational function, and the resulting expression can be algebraically manipulated into multiple equivalent forms.
Let $u = 1 + \sqrt{x}$, then $\sqrt{x} = u - 1$ and $x = (u-1)^2$, so $dx = 2(u-1)du$. The integral becomes $I = \int \frac{2(u-1)}{u^6}du = 2\int\left(\frac{1}{u^5} - \frac{1}{u^6}\right)du = 2\left(\frac{-1}{4u^4} + \frac{1}{5u^5}\right) + c$. Substituting back and simplifying: $I = \frac{-1}{2(1+\sqrt{x})^4}\left(\frac{4(u-1) + 5}{5(u-1)}\right) = \frac{-1}{2(1+\sqrt{x})^4}\left(\frac{4\sqrt{x}+1}{5\sqrt{x}}\right)$. This can be rewritten as option (1) or (2) through algebraic manipulation—both forms are equivalent, making option (3) correct.
Correct Answer: 1,2,3