If the integral $I =\int\frac{x^2}{\sqrt{1+x}}dx =\frac{1}{α} (1 + x)^α-\frac{8α}{15} (1 + x)^{α -1}+2(1+ x)^{α-2}+ C$, C is constant of integration, then the value of $α$ is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5/2
SolutionLet $t=1+x \;\;\Rightarrow\;\; dt=dx,\; x=t-1$.
$\frac{x^{2}}{\sqrt{1+x}}=\frac{(t-1)^{2}}{t^{1/2}}=(t^{2}-2t+1)t^{-1/2}=t^{3/2}-2t^{1/2}+t^{-1/2}$
Integrating term by term:
$I=\int t^{3/2}dt-2\int t^{1/2}dt+\int t^{-1/2}dt$
$I=\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+2t^{1/2}+C$
Given expression: $\frac{1}{\alpha}t^{\alpha}-\frac{8\alpha}{15}t^{\alpha-1}+2t^{\alpha-2}+C$
Exponents comparison: $\alpha=\frac{5}{2},\;\alpha-1=\frac{3}{2},\;\alpha-2=\frac{1}{2}$
Coefficient check: $\frac{1}{\alpha}=\frac{2}{5}\;\Rightarrow\;\alpha=\frac{5}{2}$
Also $-\frac{8\alpha}{15}=-\frac{8\cdot 5/2}{15}=-\frac{4}{3}$, matches perfectly.
Answer$\alpha=\frac{5}{2}$