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MCQMediumJEE 2026Indefinite Integrals

JEE Mathematics 2026 Question with Solution

Let f(x)=∫(2−x2)⋅ex(1+x)(1−x)3/2 dxf(x) = \int \frac{(2 - x^2) \cdot e^x}{(\sqrt{1 + x})(1 - x)^{3/2}} \, dx. If f(0)=0f(0) = 0, then f(12)f\left(\frac{1}{2}\right) is equal to :

  • A

    2e−1\sqrt{2e} - 1

  • B

    3e−1\sqrt{3e} - 1

  • C

    3e+1\sqrt{3e} + 1

  • D

    2e+1\sqrt{2e} + 1

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

f(x)=∫(2−x2)ex(1+x)(1−x)3/2 dxf(x) = \int \frac{(2 - x^2)e^x}{(\sqrt{1 + x})(1 - x)^{3/2}} \, dx

and f(0)=0f(0) = 0.

Find: f(12)f\left(\frac{1}{2}\right).

Use the pattern

∫ex(g(x)+g′(x)) dx=exg(x)+C\int e^x \left(g(x) + g'(x)\right) \, dx = e^x g(x) + C

if the algebraic part can be expressed as g(x)+g′(x)g(x) + g'(x).

Rewrite the integrand as

I=∫ex[2−x2(1−x2)1/2(1−x)]dxI = \int e^x \left[ \frac{2 - x^2}{(1 - x^2)^{1/2}(1 - x)} \right] dx

and test

g(x)=1+x1−x.g(x) = \sqrt{\frac{1+x}{1-x}}.

Then

g′(x)=121+x1−x⋅(1−x)−(1+x)(−1)(1−x)2=1(1−x)1−x2.g'(x) = \frac{1}{2\sqrt{\frac{1+x}{1-x}}} \cdot \frac{(1-x) - (1+x)(-1)}{(1-x)^2} = \frac{1}{(1-x)\sqrt{1-x^2}}.

Now split the numerator as

2−x2=(1−x2)+1.2 - x^2 = (1 - x^2) + 1.

So the algebraic factor becomes

2−x2(1−x2)1/2(1−x)=1−x21−x+1(1−x)1−x2=g(x)+g′(x).\frac{2 - x^2}{(1 - x^2)^{1/2}(1 - x)} = \frac{\sqrt{1-x^2}}{1-x} + \frac{1}{(1-x)\sqrt{1-x^2}} = g(x) + g'(x).

Therefore,

f(x)=ex1+x1−x+C.f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C.

Using f(0)=0f(0) = 0,

e01+01−0+C=0e^0 \sqrt{\frac{1+0}{1-0}} + C = 0 1+C=01 + C = 0 C=−1.C = -1.

Hence,

f(x)=ex1+x1−x−1.f(x) = e^x \sqrt{\frac{1+x}{1-x}} - 1.

Now evaluate at x=12x = \frac{1}{2}:

f(12)=e1/21+1/21−1/2−1=e3/21/2−1=3e−1.f\left(\frac{1}{2}\right) = e^{1/2} \sqrt{\frac{1+1/2}{1-1/2}} - 1 = \sqrt{e} \sqrt{\frac{3/2}{1/2}} - 1 = \sqrt{3e} - 1.

Therefore, the correct option is B.

Recognizing the derivative pair

Given: The integrand contains exe^x multiplied by an algebraic expression.

Find: A suitable function g(x)g(x) such that the integrand becomes ex(g+g′)e^x(g+g').

A useful trial is

g(x)=1+x1−xg(x) = \sqrt{\frac{1+x}{1-x}}

because the denominator involves both 1+x\sqrt{1+x} and powers of 1−x1-x.

Observe that

g(x)=1+x1−x=1−x21−x.g(x) = \frac{\sqrt{1+x}}{\sqrt{1-x}} = \frac{\sqrt{1-x^2}}{1-x}.

Also,

g′(x)=1(1−x)1−x2.g'(x) = \frac{1}{(1-x)\sqrt{1-x^2}}.

Then

g(x)+g′(x)=1−x21−x+1(1−x)1−x2=(1−x2)+1(1−x)1−x2=2−x2(1−x)1−x2.g(x) + g'(x) = \frac{\sqrt{1-x^2}}{1-x} + \frac{1}{(1-x)\sqrt{1-x^2}} = \frac{(1-x^2)+1}{(1-x)\sqrt{1-x^2}} = \frac{2-x^2}{(1-x)\sqrt{1-x^2}}.

This matches the integrand exactly.

Hence,

∫ex(g+g′) dx=exg(x)+C=ex1+x1−x+C.\int e^x(g+g') \, dx = e^x g(x) + C = e^x \sqrt{\frac{1+x}{1-x}} + C.

Applying the condition f(0)=0f(0)=0 gives C=−1C=-1, and then substituting x=12x=\frac12 gives f(12)=3e−1f\left(\frac12\right)=\sqrt{3e}-1. So the correct option is B.

Common mistakes

  • Choosing an arbitrary substitution for the integral without first checking the standard pattern ∫ex(g+g′) dx\int e^x(g+g') \, dx. This is inefficient because the presence of exe^x often signals a derivative-pair structure. Instead, identify a likely algebraic function g(x)g(x) and verify whether the remaining factor equals g(x)+g′(x)g(x)+g'(x).

  • Differentiating g(x)=1+x1−xg(x)=\sqrt{\frac{1+x}{1-x}} incorrectly. This leads to a mismatch with the integrand. Use the chain rule carefully and simplify to g′(x)=1(1−x)1−x2g'(x)=\frac{1}{(1-x)\sqrt{1-x^2}}.

  • Using the condition f(0)=0f(0)=0 incorrectly by forgetting the constant of integration. Since the antiderivative is ex1+x1−x+Ce^x\sqrt{\frac{1+x}{1-x}}+C, substitute x=0x=0 to determine CC before evaluating f(12)f\left(\frac12\right).

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