Intergreeting integrals #2

Calculus Level 3

Find the value of ∫ 1 e 1 + x 2 ln ⁡ x x + x 2 ln ⁡ x d x . \int_{1}^{e}\dfrac{1+x^2 \ln x}{x+x^2 \ln x}dx.

e -ln(1+e) e +ln(1+e) e ln(1+e)

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Parag Zode by Parag Zode , 24, India

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1 solution

Parag Zode
Dec 2, 2014

I = ∫ 1 e 1 + x 2 l n x x + x 2 l n x I=\int_{1}^{e}\dfrac{1+x^2lnx}{x+x^2lnx} I = ∫ 1 e ( 1 x + 1 ) d x − ∫ 1 e 1 + l n x 1 + x l n x . d x I=\int_{1}^{e}(\dfrac{1}{x}+1)dx-\int_{1}^{e}\dfrac{1+lnx}{1+xlnx}.dx I = ( l n x + x ) 1 e − ( 1 + l n x ) 1 e I=\displaystyle(lnx+x)_{1}^{e}-(1+lnx)_{1}^{e} I = e − l n ( 1 + e ) I=e-ln(1+e) is the required answer...

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