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Indefinite Integrals

Section: Mathematics

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Question : 6 of 90

Marks: +1, -0

Let

f(x)=\int \frac{(2-x^2) \cdot e^x}{(\sqrt{1+x})(1-x)^{\frac{3}{2}}} dx

f(0)=0

f\left(\,\frac{1}{2}\right)

[23 Jan 2026 Shift 1]

$\sqrt{2e}-1$
$\sqrt{2e}+1$
$\sqrt{3e}-1$
$\sqrt{3e}+1$

Solution:

Concept:

The integrand is of the form

e^x[g(x)+g'(x)]

, which integrates to

e^x g(x) + C

.

Formula:

\int e^{x}[g(x)+g'(x)]\,dx = e^{x}g(x)+C

Solution:

Rewrite denominator:

\sqrt{1+x}\,(1-x)^{3/2} = \sqrt{1-x^2}\,(1-x)

.

So integrand becomes

\frac{(2-x^{2})e^{x}}{\sqrt{1-x^2}\,(1-x)}

.

Split numerator:

2-x^2 = 1+(1-x^2)

.

Thus integral:

\int e^{x}\left[\frac{1}{\sqrt{1-x^2}\,(1-x)}+\frac{1+x}{\sqrt{1-x^2}}\right]dx

.

Let

g(x)=\frac{1+x}{\sqrt{1-x^2}}

.

Compute

g'(x)

: using quotient rule,

g'(x)=\frac{1}{\sqrt{1-x^2}(1-x)}

.

Thus the integral is

\int e^{x}[g(x)+g'(x)]dx = e^{x}g(x)+C

.

So

f(x)=\frac{e^{x}(1+x)}{\sqrt{1-x^2}}+C

.

Given

f(0)=0

:

0=\frac{1\cdot1}{1}+C ⇒ C=-1

.

Thus

f(x)=\frac{e^{x}(1+x)}{\sqrt{1-x^2}}-1

.

Evaluate at

x=\frac12

:

f\left(\frac12\right)=\frac{e^{1/2}(1+\frac12)}{\sqrt{1-(\frac12)^2}}-1 = \frac{\sqrt{e}\cdot\frac32}{\sqrt{1-\frac14}}-1 = \frac{\frac{3\sqrt{e}}{2}}{\frac{\sqrt3}{2}}-1 = \frac{3\sqrt{e}}{\sqrt3}-1 = \sqrt{3e}-1

.

Answer:

\sqrt{3e}-1

(Option C).

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