If the integral \(I = \int\displaystyle\frac{x^2}{\sqrt{1+x}}\,dx = \displaystyle\frac{1}{\alpha}(1+x)^{\alpha} - \displaystyle\frac{8\alpha}{15}(1+x)^{\alpha-1} + 2(1+x)^{\alpha-2} + C\), C is constant of integration, then the value of \(\alpha\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5}{2}\)
Let \(t = 1 + x\). Then \(dx = dt\) and \(x = t - 1\), so the integrand becomes a sum of powers of t:
\( \displaystyle \frac{x^2}{\sqrt{1+x}} = \frac{(t-1)^2}{t^{1/2}} = \frac{t^2 - 2t + 1}{t^{1/2}} \)
\( = t^{3/2} - 2t^{1/2} + t^{-1/2} \)
Integrating term by term:
\( \displaystyle I = \frac{2}{5}t^{5/2} - \frac{4}{3}t^{3/2} + 2t^{1/2} + C \)
\( \displaystyle = \frac{2}{5}(1+x)^{5/2} - \frac{4}{3}(1+x)^{3/2} + 2(1+x)^{1/2} + C \)
Comparing with the given form, the highest power gives \(\alpha = \frac{5}{2}\). The other terms agree:
\( \displaystyle \frac{1}{\alpha} = \frac{2}{5} \)
\( \displaystyle \frac{8\alpha}{15} = \frac{8}{15} \times \frac{5}{2} = \frac{4}{3} \)
\( \displaystyle \alpha - 1 = \frac{3}{2},\quad \alpha - 2 = \frac{1}{2} \)