Mathematics

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

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


ANSWER

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
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