Indefinite IntegrationeasyFree

Integral of eˣ(x−1)(x−lnx)/x² | JEE

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

Question

\displaystyle\int \dfrac{e^{x}(x-1)(x-\ln x)}{x^{2}}\,dx

is equal to

e^{x}\left(\dfrac{x-\ln x}{x}\right)+c

e^{x}\left(\dfrac{x-\ln x+1}{x}\right)+c

e^{x}\left(\dfrac{\ln x-x}{x}\right)+c

e^{x}\left(\dfrac{x-\ln x-1}{x}\right)+c

correct

Solution

Step 1: Substitute

t=\dfrac{e^{x}}{x}

. Differentiating by the quotient rule,

dt=\dfrac{e^{x}\cdot x-e^{x}\cdot 1}{x^{2}}\,dx=\dfrac{e^{x}(x-1)}{x^{2}}\,dx

which is one factor of the integrand. Step 2: Express the remaining factor through

t

. Taking logarithms of

t=\dfrac{e^{x}}{x}

\ln t=x-\ln x

so the factor

(x-\ln x)

equals

\ln t

. Step 3: Rewrite the whole integral:

I=\int (x-\ln x)\cdot\dfrac{e^{x}(x-1)}{x^{2}}\,dx=\int \ln t\,dt

Step 4: Integrate by parts:

\displaystyle\int \ln t\,dt=t\ln t-t+c

. Step 5: Back-substitute

t=\dfrac{e^{x}}{x}

and

\ln t=x-\ln x

I=\dfrac{e^{x}}{x}(x-\ln x)-\dfrac{e^{x}}{x}+c=\dfrac{e^{x}\big[(x-\ln x)-1\big]}{x}+c=e^{x}\left(\dfrac{x-\ln x-1}{x}\right)+c

Correct answer: (4)

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