Mathematics

$I$ : $\displaystyle \int e^{x}(1-\cot x+\cot^{2}x)dx=-e^{x}\cot x+c$$\displaystyle II:\int e^{x}(\frac{1+x\log x}{x})dx=e^{x}\log x+c$

both I and II are true

SOLUTION
$SI\displaystyle =\int e^x(cosec^2x-cot x)dx$
$\displaystyle =-\int e^x(cot x-cosec^2 x)dx$
$\displaystyle =-e^x cot x+c$
$SII\displaystyle =\int e^x(\frac {1}{x}+log x)dx=e^x log x+c$
Both are in the form of
$\displaystyle \int e^x(f(x)+f^1(x)]dx=e^xf(x)+c$
both are true

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

Realted Questions

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