Question

the integral of ln(2x + 1)dx, using integration by parts. I don't know where to start, please help.

Answer 1

Make the u equal to the ln(2x+1) and the dv = dx

Answer 2

http://www.wolframalpha.com/input/?i=integrate+ln%282x+%2B+1%29

Answer 3

I did that, and was stuck at the \[-2\int\limits(x/2x+1) dx\]

Answer 4

You gotta suppose u=2x+1

Answer 5

that won't work using substitution
you have to use an integral table for the integral of x/(2x+1)
it is (x/2)-(1/4)(ln|2x+1|

Answer 6

this comes up so frequently, especially in exams, that it is worth your while to remember that
\[\int \ln(x)dx=x\ln(x)-x\]

Answer 7

I used u sub like roshan2004 said, and the integrated by parts the ln u.

Answer 8

this is arrived at by integration by parts, but it is very very common. so just remember it, like remembering that
\[8\times 7=56\]
or that
\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\]

Answer 9

then when you see
\[\int \ln(2x+1)dx\] it is now a simple u - sub. put
\[u=2x+1, dx =\frac{1}{2}du\] and get
\[\frac{1}{2}\int \ln(u)du=\frac{1}{2}(u \ln(u)-u)\]
\[=\frac{1}{2}((2x+1)\ln(2x+1)-(2x+1))\]

Answer 10

you can of course integrate by parts again starting at
\[\int\ln(u)du\] but you will get
\[u\ln(u)-u\] every time, so why waste your time? just remember it

Answer 11

thank you. I will memorize the integral of ln x. thanks for throwing in the derivative of square root of x too.