Question

ind_int ln(2x+1)dx

u=ln(2x+1) du=2/(2x+1)dx
dv=dx v=x

Answer 1

now plug them in

Answer 2

uv-int(v du)

Answer 3

i got as far as \[\int\limits_{}^{} xln(2x+1)-2\int\limits_{}^{}x/2x+1 dx\]

Answer 4

I'm stuck.

Answer 5

you have an extra integral in there

Answer 6

hmm for some reason I the equation editor is showing me the code not the symbels

Answer 7

refresh @zzr0ck3r

Answer 8

ahh ty

Answer 9

welcome^\[\int\ln(2x+1)dx\]\[u=\ln(2x+1)\implies du={2dx\over2x+1}\]\[dv=dx\implies v=x\]\[\int udv=uv-\int vdu=x\ln(2x+1)-\int{2x\over2x+1}dx\]now use long division on the integral

Answer 10

I think my problem lies in remembering how to integrate x/2x+1...quotient rule?

Answer 11

no like I say, do polynomial long division on\[\frac{2x}{2x+1}\]if you remember how from probably a few years back, hehe...

Answer 12

ok, let me try, and I'll type my answer. It'll probably be a little while :-)

Answer 13

you can use sub method or uv method for that also

Answer 14

thanks zz, and tt, I'll try both

Answer 15

The book says:
\[xln(2x+1)-\int\limits_{}^{}2x/(2x+1)dx=xln(2x+1)-\int\limits_{}^{}(2x+1)-1/(2x+1)dx\]

Answer 16

I think I'm forgetting algebra because I don't understand 2x/2x+1=(2x+1)-1/(2x+1)

Answer 17

lol