Question

evaluate ∫x ln5 xdx using integration by parts. Use U=ln 5x . anyonce can help me?

Answer 1

\[\Huge \int\limits xln(5x)dx?\]

Answer 2

yes @.Sam.

Answer 3

integration by parts?

Answer 4

Use U=ln 5x. @AravindG

Answer 5

Oh sorry ..I didnt read that

Answer 6

First we'll do u-sub
\[ u=5x ~~~~~~~\frac{u}{5}=x\\ \\ du=5dx \\ \\ \frac{du}{5}=dx \\ \\ \frac{1}{25} \int\limits uln(u)du\]

Answer 7

IBP\[Let \\ \\ J=\ln(u)~~~~~dk=udu \\ \\ dJ=\frac{1}{u}du ~~~~~~k=\frac{u^2}{2} \\ \\ \frac{1}{25}[\frac{u^2}{2}\ln(u)-\int\limits \frac{u^{\cancel 2}}{2}\frac{1}{\cancel u}du ]\\ \\ \frac{1}{25}[\frac{u^2}{2}\ln(u)-\frac{u^2}{4}] \\ \\ \frac{1}{25}[\frac{25x^2}{2}\ln(5x)-\frac{25x^2}{4}] \\ \\ \frac{x^2}{2}\ln(5x)-\frac{x^2}{4}\]

Answer 8

Oh and don't forget the constant

Answer 9

The formula for J and K integration by parts here is
\[\int\limits xydx \\ \\ \int\limits \overbrace{x}^{J} \overbrace{ydy}^{dK} \\ \text{Find dJ and K then substitute into} \\ JK-\int\limits K~dJ\]

Answer 10

how to find the dJ and k ? @.Sam.

Answer 11

For your question, we set \[J=\ln(u) ~~~ \text{and} ~~~ dK=udu \\ \\ \]
So differentiating J, we get dJ, and integrating dK, we get K
\[ dJ=\frac{1}{u}du ~~~~~~k=\frac{u^2}{2}\]

Answer 12

Then substitute into this\[JK-\int\limits\limits K~dJ\]
and get
\[ \frac{1}{25}[\frac{u^2}{2}\ln(u)-\int\limits\limits \frac{u^{ 2}}{2}\frac{1}{ u}du ]\]

Answer 13

there is a formula to solve such types of problems that contains two terms one you can integrate easily and second that you can't. so in order to solve such problems the formula is
(first term) cross (integration of second term)-integration{(differentiation of first term)-(integration of second term)dx}dx
firs term=that you can't integrate easily
second term= that you can integrate easily
your first term is ln(5x) and second term is x.