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Integral of xˣ(1/x + ln²x + lnx) | JEE

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

Question
∫xx ⁣(1x+ln⁡2x+ln⁡x)dx\displaystyle\int x^{x}\!\left(\dfrac{1}{x}+\ln^{2}x+\ln x\right)dx equals
Axx ⁣(ln⁡2x−1x)+Cx^{x}\!\left(\ln^{2}x-\dfrac{1}{x}\right)+C
Bxx(ln⁡x−x)x^{x}(\ln x-x)
Cxxln⁡2x2+C\dfrac{x^{x}\ln^{2}x}{2}+C
Dxxln⁡x+Cx^{x}\ln x+Ccorrect
Solution
Step 1: Recall ddx(xx)=xx(ln⁡x+1)\dfrac{d}{dx}(x^{x})=x^{x}(\ln x+1), obtained from xx=exln⁡xx^{x}=e^{x\ln x}. Step 2: Differentiate the candidate xxln⁡xx^{x}\ln x by the product rule:
ddx(xxln⁡x)=[xx(ln⁡x+1)]ln⁡x+xx⋅1x\dfrac{d}{dx}\big(x^{x}\ln x\big)=\Big[x^{x}(\ln x+1)\Big]\ln x+x^{x}\cdot\dfrac{1}{x}
 Step 3: Expand:
=xx(ln⁡2x+ln⁡x)+xxx=xx ⁣(1x+ln⁡2x+ln⁡x)=x^{x}(\ln^{2}x+\ln x)+\dfrac{x^{x}}{x}=x^{x}\!\left(\dfrac{1}{x}+\ln^{2}x+\ln x\right)
 which is exactly the integrand. Step 4: Therefore
I=∫xx ⁣(1x+ln⁡2x+ln⁡x)dx=xxln⁡x+CI=\int x^{x}\!\left(\dfrac{1}{x}+\ln^{2}x+\ln x\right)dx=x^{x}\ln x+C
 Correct answer: (4)
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