int frac{x+100}{(x+101)^2} e^x , dx = . . . | vedclass

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Question

(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$. Given the integral: $I = \int \frac{x+100}{(x+101)^2} e^x \,

Vedclass · Accepted Answer

$\frac{1}{x+101} e^x$

Vedclass · Answer

(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$. Given the integral: $I = \int \frac{x+100}{(x+101)^2} e^x \, dx$. We can rewrite the numerator as: $x + 100 = (x + 101) - 1$. So,the integral becomes: $I = \int \frac{(x+101) - 1}{(x+101)^2} e^x \, dx$. $I = \int \left( \frac{x+101}{(x+101)^2} - \frac{1}{(x+101)^2} \right) e^x \, dx$. $I = \int \left( \frac{1}{x+101} - \frac{1}{(x+101)^2} \right) e^x \, dx$. Let $f(x) = \frac{1}{x+101}$. Then $f'(x) = -\frac{1}{(x+101)^2}$. Since the expression is in the form $\int e^x [f(x) + f'(x)] \, dx$,the result is $e^x f(x) + C$. Therefore,$I = e^x \left( \frac{1}{x+101} \right) + C = \frac{e^x}{x+101} + C$.

$\int \frac{x+100}{(x+101)^2} e^x \, dx = $ . . . . . . $+ C$.

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