Question

help with integration by parts ?

Answer 1

What kinda help?

Answer 2

\[\int\limits(\ln5x)/x^2 dx\]

Answer 3

Maybe try to differentiate the ln(5x) part
and integrate the 1/x^2 part

Answer 4

\[\int\limits_{}^{}\ln(5x) \cdot x^{-2} dx \]
Choose u=ln(5x) and dv=x^(-2)

Answer 5

oh okay ! thanks!

Answer 6

Did you want to try it before you say thanks? :p

Answer 7

lol i meant for the idea cause i started with u=x^2 and got stuck lol

Answer 8

ah ok lol

Answer 9

so i got (ln5x)(-1/x) + ln5x^2 +c does that seem right lol

Answer 10

hmm half right

Answer 11

for the second term how did you get ln5x^2?

Answer 12

(-1/x)(1/5x) which was -1/5x^2 which integrated comes to -ln5x^2

Answer 13

\[\int\limits_{}^{}\ln(5x) x^{-2} dx = -x^{-1}\ln(5x)-\int\limits_{}^{}-x^{-1}\frac{5}{5x} dx \]
\[=\frac{-\ln(5x)}{x}+\int\limits_{}^{}x^{-2} dx \]

Answer 14

Use the power rule

Answer 15

i thought du = 1/5x

Answer 16

\[\frac{d}{dx}\ln(u)= u' \cdot \frac{1}{u} \text{ by chain rule } \]

Answer 17

u=5x
u'=5

Answer 18

but even if you had 1/(5x) you are still using the wrong method to evaluate
\[\int\limits_{}^{}\frac{1}{cx^2} dx=\int\limits_{}^{}\frac{1}{c} x^{-2} dx=\frac{1}{c} \frac{x^{-1}}{-1}+K=\frac{-1}{cx}+K\]

Answer 19

but yeah c should be 1 in this case since 5/5=1

Answer 20

also if you don't like that rule I just used
how about his:
\[\frac{d}{dx}\ln(cx)=\frac{d}{dx}(\ln(c)+\ln(x))=0+\frac{1}{x}\]

Answer 21

I chose to use the product rule for log expressions
and then take the derivative

Answer 22

Okay I see where I went wrong I had to use the chain rule for ln5x that's why I had the wrong value, now I got your answer thanks!!!

Answer 23

coolness

Answer 24

random question but how do you integrate sin3x dx ?