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Mathematics 52 Online
OpenStudy (anonymous):

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

OpenStudy (zzr0ck3r):

now plug them in

OpenStudy (zzr0ck3r):

uv-int(v du)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I'm stuck.

OpenStudy (turingtest):

you have an extra integral in there

OpenStudy (zzr0ck3r):

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

OpenStudy (turingtest):

refresh @zzr0ck3r

OpenStudy (zzr0ck3r):

ahh ty

OpenStudy (turingtest):

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

OpenStudy (anonymous):

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

OpenStudy (turingtest):

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

OpenStudy (anonymous):

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

OpenStudy (zzr0ck3r):

you can use sub method or uv method for that also

OpenStudy (anonymous):

thanks zz, and tt, I'll try both

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

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