Question

\[\int\limits_{}^{}\ln(2x+1)dx = xln(2x+1) - \int\limits_{}^{} 2dx/(2x+1) \]
u = 2x+1
du = 2dx

thus

\[xln(2x+1) - \int\limits_{}^{}du/u = xln(2x+1) - \ln|2x+1| + c\]

Is this correct?

Answer 1

not quite, I think we need partial fractions for the second part
you dropped an x

Answer 2

oh yeah you are right I did

Answer 3

should be

\[xln(2x+1) - \int\limits_{}^{} 2xdx/(2x + 1)\]

Answer 4

better yet, write it as\[\int\ln(2x+1)dx = xln(2x+1) - \int \frac{2xdx}{2x+1}\]however you want to do this
I suggest a u-sub\[u=2x+1\implies2x=u-1\]

Answer 5

u = 2x + 1
du = 2dx

thus,

x ln(2x+1) - \[\int\limits_{}^{} dux/u\]

Answer 6

but yeah that isn't right :\

Answer 7

oh nvm :)

\[\ln(2x+1)x - \int\limits_{}^{} ((1-u/2)/u)du\]

Answer 8

yeah that looks better

Answer 9

\[u=2x+1\implies2x=u-1\]\[du=2dx\]so we have the second integral as\[\frac12\int1-u^{-1}du\]I think you dropped a 1/2, or at least forgot to put it outside the parentheses

Answer 10

=

\[\ln(2x+1)x - \int\limits_{}^{} ((1-u)/2u)du = \ln(2x+1)x - \int\limits_{}^{}(1/2u)du - \int\limits_{}^{}(u/2u)du\]

ok I think I got it

Answer 11

yeah you are right i did

Answer 12

simplify, and be careful with your signs;
you forgot to distribute a negative\[\ln(2x+1)x - \int\limits_{}^{} ((1-u)/2u)du = \ln(2x+1)x - \int\limits_{}^{}(1/2u)du + \int(\cancel{u}/2\cancel{u}^1)du\]\[\ln(2x+1)x - \int\limits_{}^{} ((1-u)/2u)du = \ln(2x+1)x - \int\limits_{}^{}(1/2u)du -+\int\limits_{}^{}(1/2)du\]

Answer 13

\[x\ln(2x+1)-\frac12\int1-u^{-1}du\]is the way I prefer to see it :)

Answer 14

the answer being

xln(2x+1) +x/2 -(1/4)ln|2x+1| + c

Answer 15

crud I made a mistake :l

Answer 16

\[\frac12\int1-u^{-1}du=\frac12u-\frac12\ln|u|+C=\frac12(2x+1)-\frac12\ln(2x+1)+C\]so we have\[x\ln(2x+1)-\frac12(2x+1)+\frac12\ln(2x+1)+C\]which you can simplify

Answer 17

\[\ln(2x+1)(x+\frac12)-x-\frac12+C\]since 1/2 is a constant we can absorb it inot C and get\[\ln(2x+1)(x+\frac12)-x+C\]

Answer 18

I will try it agin Thanks for the help

Answer 19

welcome :)