Question

Integration by parts:

the integral of ln(2x + 1) dx

I am stuck after trying the formula. please help me.

Answer 1

Let Int denote the integral sign. Integration by parts is

Int[u(dv)] = u*v - Int[v(du)]

Let u =ln( 2x + 1) so du = [2(dx)]/[2x+1]
dv = dx so v = x

uv = x*ln(2x+1)

Int[v(du)] = Int[(2x)/(2x+1)](dx)

Since 2x/(2x+1) = 1 - 1/(2x+1) we split it up into

Int[(dx)] - Int[(1/(2x+1))(dx)]

Int[(dx)] = x

Int[(1/(2x+1))(dx)] = (ln(2x+1))/2

The integral is then

x*ln(2x+1) - x + ln(2x+1)/2

Which can be verified by taking the derivative

Answer 2

You want to start off with a substitution. Let p=2x+1
then dp=2dx
Rewrite the integral this way:
\[\frac{1}{2}\int\limits \ln(p)dp\]
From here, let u=ln(p) and dv=dp
so du=1/p and v=p
Using the formula you have:
\[\frac{1}{2}(p \ln(p)-\int\limits \frac{p}{p}dx)\]
This reduces to:\[\frac{1}{2}(p \ln(p)-\int\limits dp)=\frac{1}{2}(p \ln(p)-p)\]
Now replug in your p:
\[\frac{1}{2}(2x+1)\ln(2x+1)-\frac{1}{2}(2x+1)+C\]
Does that make sense?

Answer 3

above answer is absolutely right. but if i were you (taking a calc course) i would just try to remember that
\[\int\ln(x)dx=x\ln(x)-x\] then if you know that the above is a simple u-sub. integrating the natural log is a question math teachers love to ask (parts is the way) so i would just memorize it.