JEE-Mains 2014 (14/30)

Calculus Level 3

∫ ( 1 + x − 1 x ) e x + 1 x d x = ? \large \int \left( 1+x-\frac{1}{x} \right) e^{x+\frac{1}{x}}\, dx = \ ?

( x − 1 ) e x + 1 x + c (x-1)e^{x+\frac{1}{x}}+c − x e x + 1 x + c -xe^{x+\frac{1}{x}}+c x e x + 1 x + c xe^{x+\frac{1}{x}}+c ( x + 1 ) e x + 1 x + c (x+1)e^{x+\frac{1}{x}}+c

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Akshat Sharda
Jul 29, 2017

∫ { e x + 1 x + x e x + 1 x ( 1 − 1 x 2 ) } d x = x e x + 1 x + C ∵ ∫ { f ( x ) + x f ′ ( x ) } d x = x f ( x ) + C \int \left\{ e^{x+\frac{1}{x}}+xe^{x+\frac{1}{x}}\left(1-\frac{1}{x^2}\right) \right\} dx = xe^{x+\frac{1}{x}} +C \\ \because \int \left\{ f(x)+xf'(x)\right\} dx=xf(x)+C

11 Helpful 0 Interesting 2 Brilliant 0 Confused

Simple and elegant :D

John Frank - 3 years, 9 months ago

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