Question

What is the integral of `(lnx+x^2e^x)/x`?

Answer 1

`int(ln+x^2e^x)/xdx=(x-1)e^x+1/2ln^2x+"c"`

Explanation:

`int(lnx+x^2e^x)/xdx=intlnx/x+xe^xdx=intlnx/x+intxe^xdx`

For `intln/xdx`, let `u=lnx` and `du=1/xdx`

So

`intln/xdx=intudu=1/2u^2=1/2ln^2x +"c"_1`

For `intxe^xdx`, integrate by parts.

Let `u=xrArrdu=dx`

And `dv=e^xrArrv=e^x`

So

`intxe^xdx=xe^x-inte^xdx=xe^x-e^x+"c"=(x-1)e^x+"c"_2`

Combining the two gives the final answer as

`(x-1)e^x+1/2ln^2x+"c"`