CUET · MATHS · PYQ PAPER 2025
If the integral \(I=\int \frac{x^2}{\sqrt{1+x}} d x=\frac{1}{\alpha}(1+x)^\alpha-\frac{8 \alpha}{15}(1+x)^{\alpha-1}+2(1+x)^{\alpha-2}+C\), where \(C\) is the constant of integration, then the value of \(\alpha\) is :
- A 1
- B \(\frac{1}{2}\)
- C \(\frac{3}{2}\)
- D \(\frac{5}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
Let \(u=1+x\), so \(x=u-1\) and \(dx=du\). \(I = \int \frac{(u-1)^2}{\sqrt{u}} du = \int (u^{3/2} - 2u^{1/2} + u^{-1/2}) du\) \(I = \frac{u^{5/2}}{5/2} - 2\frac{u^{3/2}}{3/2} + \frac{u^{1/2}}{1/2} + C = \frac{2}{5}(1+x)^{5/2} - \frac{4}{3}(1+x)^{3/2} + 2(1+x)^{1/2} + C\)…
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