Indefinite IntegrationmediumFree

Double Integral of eˣ(lnx + 2/x − 1/x²) | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int\!\left(\int e^{x}\!\left(\ln x+\dfrac{2}{x}-\dfrac{1}{x^{2}}\right)dx\right)dx

equals

e^{x}\ln x+C_{1}x+C_{2}

correct

e^{x}\ln x+\dfrac{1}{x}+C_{1}x+C_{2}

\dfrac{\ln x}{x}+C_{1}x+C_{2}

DNone of these

Solution

Step 1: Resolve the inner integral first. Choose

f(x)=\ln x+\dfrac{1}{x}

, whose derivative is

f'(x)=\dfrac{1}{x}-\dfrac{1}{x^{2}}

Adding

f

and

f'

f(x)+f'(x)=\left(\ln x+\dfrac{1}{x}\right)+\left(\dfrac{1}{x}-\dfrac{1}{x^{2}}\right)=\ln x+\dfrac{2}{x}-\dfrac{1}{x^{2}}

which is exactly the inner integrand. Step 2: Apply the standard result

\displaystyle\int e^{x}\big(f(x)+f'(x)\big)\,dx=e^{x}f(x)

\int e^{x}\!\left(\ln x+\dfrac{2}{x}-\dfrac{1}{x^{2}}\right)dx=e^{x}\!\left(\ln x+\dfrac{1}{x}\right)+C_{1}

Step 3: Integrate this result a second time. Write

e^{x}\!\left(\ln x+\dfrac{1}{x}\right)=e^{x}\big(g(x)+g'(x)\big)

with

g(x)=\ln x,\ g'(x)=\dfrac{1}{x}

, so the same rule gives

\int e^{x}\!\left(\ln x+\dfrac{1}{x}\right)dx=e^{x}\ln x

Step 4: The constant

C_{1}

from Step 2 integrates to

C_{1}x

\int\left[e^{x}\!\left(\ln x+\tfrac{1}{x}\right)+C_{1}\right]dx=e^{x}\ln x+C_{1}x+C_{2}

Correct answer: (1)

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