Indefinite IntegrationeasyFree

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Indefinite Integration · medium
If ∫ex 1−x6−6x2(1+x3)2 dx=ex 1−xp1+xq+c\displaystyle\int e^{x}\,\dfrac{1-x^{6}-6x^{2}}{(1+x^{3})^{2}}\,dx=e^{x}\,\dfrac{1-x^{p}}{1+x^{q}}+c, then p+qp+q equals
Indefinite Integration · medium
∫dxx−x3\displaystyle\int \dfrac{dx}{\sqrt{x}-\sqrt[3]{x}} equals
Indefinite Integration · easy
If ∫x−sin⁡xcos⁡xx2cos⁡2x dx=f(x)+c\displaystyle\int \dfrac{x-\sin x\cos x}{x^{2}\cos^{2}x}\,dx=f(x)+c, then lim⁡x→0f(x)\displaystyle\lim_{x\to 0}f(x) equals
Indefinite Integration · medium
If ∫x2e−2x dx=e−2x(ax2+bx+c)+d\displaystyle\int x^{2}e^{-2x}\,dx=e^{-2x}(ax^{2}+bx+c)+d, then

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath