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$\int (\frac{1}{x^{2}}) (\frac{1}{e^{x}}) d x$ $int(1/x^2)(1/e^x)dx$

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Answer 1 · 2018-04-05T10:13:07

$- \frac{1}{e^{x}} + e^{[1 - x]} - \frac{1}{x}$ $-1/e^x+e^[[1-x]] -1/x$

Explanation:

Let $\frac{1}{x^{2} e^{x}} = \frac{A}{e^{x}} + \frac{B}{x^{2}}$ $1/[x^2e^x]=A/e^x+B/x^2$ , multiply denominator and numerator by $[x^{2} e^{x}]$ $[x^2e^x]$ . [Method of partial fractions]

So, $1 = A \frac{x^{2} e^{x}}{e^{x}}$ $1= A[x^2e^x]/e^x$ + $B \frac{x^{2} e^{x}}{x^{2}}$ $B[x^2e^x]/x^2$ = $A x^{2} + B e^{x}$ $Ax^2 +Be^x$

Let $x = 0$ $x=0$ , therefore, $1 = A [0] + B e^{0}$ $1= A[0]+Be^0$ , ie, $[B = 1]$ $[B=1]$ ............[*]

Let $x = 1$ $x= 1$ , therefore, $1$ $1$ = $A [1]$ $A[1]$ + $B e$ $Be$ ,i.e, $A = [1 - e]$ $A= [1-e]$ ,[since $B = 1$ $B=1$

So, $\int \frac{1}{x^{2} e^{x}} d x$ $int1/[x^2e^x]dx$ = $\int [\frac{1 - e}{e^{x}}$ $int [[1-e]/e^x$ + $\frac{1}{x^{2}}] d x$ $1/x^2]dx$ = $\int [\frac{1}{e} x - \frac{e}{e^{x}} + \frac{1}{x^{2}}] d x$ $int [1/ex -e/e^x+1/x^2]dx$ ,

These terms can now be integrated separately, giving the answer above. Hope this was helpful.

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$\int (\frac{1}{x^{2}}) (\frac{1}{e^{x}}) d x$ $int(1/x^2)(1/e^x)dx$

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