Integration by parts evaluation

[delapcsoncruz]

∫xa^x

u=x
du=dx
dv=a^xdx
v=a^x/lna

= xa^x - ∫a^xdx/lna

is my solution right?
my problem now is how to integrate the expression xa^x - ∫a^xdx/lna
please help..

 

[The1337gamer]

Seems correct to me, you just need to evaluate the 2nd term, the integral, you know how to do integrate a^x as you've already done it, 1/ln(a) is just a constant.

 

[delapcsoncruz]

Seems correct to me, you just need to evaluate the 2nd term, the integral, you know how to do integrate a^x as you've already done it, 1/ln(a) is just a constant.

ok this what I've got...

=xa^x - ∫a^xdx/lna

=xa^x - 1/lna∫a^xdx

=xa^x - (1/lna) (a^x/lna)

=xa^x - a^x/2lna

is that right that (lna)(lna) = 2lna?
or it is just (lna)(lna) = (lna)(lna)

 

[The1337gamer]

It's the bottom line as:

2ln(a) = ln(a^2)

What you have is ln(a)ln(a) = (ln(a))^2.

 

[delapcsoncruz]

It's the bottom line as:

2ln(a) = ln(a^2)

What you have is ln(a)ln(a) = (ln(a))^2.

but , was my answer correct?

 

[The1337gamer]

xa^x - a^x/(2lna) isn't correct, xa^x - a^x/(ln(a))^2 is.

 

[delapcsoncruz]

xa^x - a^x/(2lna) isn't correct, xa^x - a^x/(ln(a))^2 is.

ah ok..thank you very much!