ind_int ln(2x+1)dx u=ln(2x+1) du=2/(2x+1)dx dv=dx v=x
now plug them in
uv-int(v du)
i got as far as \[\int\limits_{}^{} xln(2x+1)-2\int\limits_{}^{}x/2x+1 dx\]
I'm stuck.
you have an extra integral in there
hmm for some reason I the equation editor is showing me the code not the symbels
refresh @zzr0ck3r
ahh ty
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
I think my problem lies in remembering how to integrate x/2x+1...quotient rule?
no like I say, do polynomial long division on\[\frac{2x}{2x+1}\]if you remember how from probably a few years back, hehe...
ok, let me try, and I'll type my answer. It'll probably be a little while :-)
you can use sub method or uv method for that also
thanks zz, and tt, I'll try both
The book says: \[xln(2x+1)-\int\limits_{}^{}2x/(2x+1)dx=xln(2x+1)-\int\limits_{}^{}(2x+1)-1/(2x+1)dx\]
I think I'm forgetting algebra because I don't understand 2x/2x+1=(2x+1)-1/(2x+1)
lol
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