How do you integrate $int x*5^x$ by integration by parts method?

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Answer 1 · 2017-06-11T16:38:22

$intx*5^x$ $dx=$

$intx*5^x$ $dx$

Let $f=x$ , then $df=dx$

And $dg=5^x$ $dx$ , then $g=5^x /ln5$

$intx*5^x$ $dx=(x*5^x)/ln5-1/ln5int5^x$ $dx$

$=(x*5^x)/ln5-5^x/ln^2 5+"c"$

Taking out a factor of $1/ln^2 5$ yields:

$intx*5^x$ $dx=1/ln^2 5[(x*5^x)ln5-5^x]+"c"$

How to integrate $5^x$ $dx$

$int5^x$ $dx$

Let $u=5^x$

Then $du=5^xln5$ $dx$

And $dx=1/(5^xln5)$ $du$

$thereforeint5^x$ $dx=int5^x xx 1/(5^xln5)$ $du=1/ln5int1$ $du=u/ln5+"c"$
$=5^x/ln5+"c"$

How do you integrate int x*5^xint x*5^x by integration by parts method?