Question

ind int ln(2x+1)dx=xln(2x+1)-x+(1/2)ln(2x+1)+c

Answer 1

how do you get a final ans of (1/2)(2x+1)ln(2x+1)-x+c?

Answer 2

how did you get the first answer?

Answer 3

should be by parts, although it comes up so often it is easier just to remember that
\[\int \ln(u)du=u\ln(u)-u+C\]

Answer 4

too long to type out, I tried before....but I can't understand how to get from my answer to the final ans.

Answer 5

ok let me see what we get using \(u=2x+1\) and \(\frac{1}{2}du =dx\)

Answer 6

we can go right to the answer and get \[\frac{1}{2}\left(u\ln(u)-u\right)\]

Answer 7

\[\frac{1}{2}\left((2x+1)\ln(2x+1)-(2x+1)\right)\]

Answer 8

then multiply the second term by \(\frac{1}{2}\) to get
\[\frac{1}{2}(2x+1)\ln(2x+1)-x-\frac{1}{2}\]

Answer 9

oh i see what was bothering you
the \(\frac{1}{2}\) at the end is unimportant, it is a constant

Answer 10

it gets sucked up in to the \(+C\)

Answer 11

i mean i guess that is what was bothering you, because if you do the algebra you do not get exactly what is written, but you are off by a constant which you always are