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MCQMediumJEE 2023Integration Techniques (Substitution, Parts, Partial Fractions)

JEE Mathematics 2023 Question with Solution

Let I(x)=∫0xx+1x(1+xex)2 dxI(x) = \int_{0}^{x} \frac{x+1}{x(1+x e^x)^2} \, dx, x>0x > 0. If lim⁡x→∞I(x)=0\lim_{x \to \infty} I(x) = 0, then I(1)I(1) is equal to:

  • A

    e+2e+1−log⁡(e+1)\frac{e+2}{e+1} - \log(e+1)

  • B

    e+1e+2+log⁡(e+1)\frac{e+1}{e+2} + \log(e+1)

  • C

    e+2e+1−log⁡(e+1)\frac{e+2}{e+1} - \log(e+1)

  • D

    e+1e+2+log⁡(e+1)\frac{e+1}{e+2} + \log(e+1)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: I(x)=∫0xx+1x(1+xex)2 dxI(x) = \int_{0}^{x} \frac{x+1}{x(1+x e^x)^2} \, dx with x>0x > 0 and lim⁡x→∞I(x)=0\lim_{x \to \infty} I(x) = 0.

Find: I(1)I(1).

Use the substitution shown in the solution:

1+xex=t1 + x e^x = t

Then

(xex+ex) dx=dt(x e^x + e^x) \, dx = dt

so

(x+1) dx=dtex=dtt−1.(x+1) \, dx = \frac{dt}{e^x} = \frac{dt}{t-1}.

Now simplify the integral:

I(x)=∫dtt2(t−1)=∫[1t−1−1t−1t2] dt.I(x) = \int \frac{dt}{t^2(t-1)} = \int \left[\frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2}\right] \, dt.

Integrating term by term,

I(x)=ln⁡∣t−1∣−ln⁡∣t∣+1t.I(x) = \ln|t-1| - \ln|t| + \frac{1}{t}.

At x=1x=1, we have

t=1+e.t = 1 + e.

Therefore,

I(1)=ln⁡(e1+e)+11+e.I(1) = \ln\left(\frac{e}{1+e}\right) + \frac{1}{1+e}.

Now simplify:

I(1)=ln⁡(e)−ln⁡(1+e)+11+e=1−ln⁡(1+e)+11+e.I(1) = \ln(e) - \ln(1+e) + \frac{1}{1+e} = 1 - \ln(1+e) + \frac{1}{1+e}.

Hence,

I(1)=e+2e+1−log⁡(e+1).I(1) = \frac{e+2}{e+1} - \log(e+1).

Therefore, the value of I(1)I(1) is e+2e+1−log⁡(e+1)\frac{e+2}{e+1} - \log(e+1), so the correct option is C.

Substitution and simplification

Given: I(x)=∫0xx+1x(1+xex)2 dxI(x) = \int_{0}^{x} \frac{x+1}{x(1+x e^x)^2} \, dx.

Find: I(1)I(1).

The key idea is to convert the denominator 1+xex1 + x e^x into a single variable. Take

t=1+xex.t = 1 + x e^x.

Then

dt=(xex+ex) dx=ex(x+1) dx.dt = (x e^x + e^x) \, dx = e^x(x+1) \, dx.

Also, from t=1+xext = 1 + x e^x, we get

xex=t−1.x e^x = t - 1.

Hence

exx=t−1.e^x x = t-1.

Using this together with dt=ex(x+1) dxdt = e^x(x+1) \, dx gives the transformed integrand used in the solution.

So the integral becomes

I(x)=∫dtt2(t−1).I(x) = \int \frac{dt}{t^2(t-1)}.

Now decompose:

1t2(t−1)=1t−1−1t−1t2.\frac{1}{t^2(t-1)} = \frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2}.

Integrating,

I(x)=∫(1t−1−1t−1t2) dt=ln⁡∣t−1∣−ln⁡∣t∣+1t.I(x) = \int \left(\frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2}\right) \, dt = \ln|t-1| - \ln|t| + \frac{1}{t}.

For x=1x=1,

t=1+1⋅e=1+e.t = 1 + 1\cdot e = 1+e.

Thus,

I(1)=ln⁡(e1+e)+11+e.I(1) = \ln\left(\frac{e}{1+e}\right) + \frac{1}{1+e}.

Finally,

ln⁡(e1+e)=ln⁡(e)−ln⁡(1+e)=1−ln⁡(1+e).\ln\left(\frac{e}{1+e}\right) = \ln(e) - \ln(1+e) = 1 - \ln(1+e).

Therefore,

I(1)=1−ln⁡(1+e)+11+e=e+2e+1−log⁡(e+1).I(1) = 1 - \ln(1+e) + \frac{1}{1+e} = \frac{e+2}{e+1} - \log(e+1).

Therefore, the correct option is C.

Common mistakes

  • Using the same variable xx as both the upper limit and the dummy variable of integration without recognizing the substitution step. This causes confusion in transforming the integrand. Treat the integrand carefully under the substitution t=1+xext = 1 + x e^x as shown in the solution.

  • Differentiating 1+xex1 + x e^x incorrectly. The derivative is ex(x+1)e^x(x+1), not only xexx e^x or only exe^x. Use the product rule correctly before replacing terms in the integral.

  • Making an error in partial fraction decomposition of 1t2(t−1)\frac{1}{t^2(t-1)}. If the decomposition is wrong, the logarithmic and rational terms will be incorrect. Verify that 1t2(t−1)=1t−1−1t−1t2\frac{1}{t^2(t-1)} = \frac{1}{t-1} - \frac{1}{t} - \frac{1}{t^2} before integrating.

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