Question

How do you integrate

x^(7/2) ln x dx ?

Answer 1

\[\int\limits_{}^{}x^{7/2}lnx dx\]

Answer 2

u= x^7/2
dv=lnx
v=? idk lol i'm stuck
du= 7/2x^5/2

Answer 3

\[\int\limits x ^{\frac{ 7 }{ 2 }}\ln x~dx=\ln x \frac{ x ^{\frac{ 9 }{ 2 }} }{ \frac{ 9 }{ 2 } }-\int\limits \frac{ 1 }{ x }* \frac{ x ^{\frac{ 9 }{ 2 }} }{ {\frac{ 9 }{ 2 }} }dx\]
=?

Answer 4

integration by parts

Answer 5

ln x is first function

Answer 6

ya you've got your parts backwards :)
remember that when you differentiate natural log, it turns into some of that x business.

So you want your ln x = u

Answer 7

\[\int\limits uv dx=u \int\limits vdx-u' \int\limits vdx\]

Answer 8

"u= x^7/2
dv=lnx
v=? idk lol i'm stuck"

^Right there you should have known that you chose u and dv incorrectly.