How to integrate $intx^2e^(-5x)dx$ by using integration by parts?

Question

$intx^2e^(-5x)dx$

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Answer 1 · 2018-04-30T13:44:18

$I=-e^(-5x)[x^2/5+(2x)/25+2/125]+c$

Here,

$I=intx^2e^(-5x)dx$

$"Using "color(blue) "Integration by Parts":$

$int(u*v)dx=uintvdx-int(u'intvdx)dx$

Let, $u=x^2 and v=e^(-5x)$

$=>u'=2x and intvdx=(e^(-5x))/(_5)$

$I=x^2(e^(-5x)/(-5))-int2x(e^(-5x)/(-5))dx$

$=-x^2/5e^(-5x)+2/5intxe^(-5x)dx$

Again, $"Using "color(blue) "Integration by Parts":$ in second
integral

$=-x^2/5e^(-5x)+2/5[x(e^(_5x)/(-5))-int(e^(-5x)/(-5))dx]$

$=-x^2/5e^(-5x)+2/5[-x/5e^(-5x)+1/5(e^(-5x)/(-5))]+c$

$=-x^2/5e^(-5x)-(2x)/25e^(-5x)-2/125e^(-5x)+c$

$I=-e^(-5x)[x^2/5+(2x)/25+2/125]+c$