Evaluate the integral: ln(2x + 1)dx

[MarkSA]

Hello,

1) Evaluate the integral of: ln(2x + 1)dx

I'm supposed to do this with integration by parts. The problem is, when I do that, with u = ln(2x + 1), I get:
= xln(2x + 1) - integral of: [2x/(2x + 1)]dx. I can't figure out what to do with the result.

A solutions manual showed that to solve it, they did:
Integral of [(2x + 1) - 1]/(2x + 1). They then broke it up into two parts to solve it. This seems like a fancy trick though (that I would probably not think of on the spot), and i'm wondering if there must be another easier way to solve it? This is our very first section on integration by parts and one of the first questions so this should only involve the basics...

Any ideas?

 

[galactus]

Why couldn't you just let \(\displaystyle u=2x+1, \;\ \frac{du}{2}=dx\)?.

This gives:

\(\displaystyle \frac{1}{2}\int{ln(u)}du\)

Now, integrate and resub.

EDIT: Oh, I am sorry. You must use parts.

Let \(\displaystyle u=ln(2x+1), \;\ dv=dx, \;\ du=\frac{2x}{2x+1}dx, \;\ v=x\)

Now, put it together and it should work OK.

 

[soroban]

Hello, Mark!

\(\displaystyle \text{1) Evaluate the integral: }\int \ln(2x + 1)\,dx\)

\(\displaystyle \text{I'm supposed to do this with integration by parts.}\)
\(\displaystyle \text{The problem is, when I do that, with }\:u \:= \:\ln(2x + 1),\)
\(\displaystyle \text{I get: }\;x\ln(2x + 1) - \int\frac{2x}{2x + 1}\,dx\)

I can't figure out what to do with the result.

Try long division . . .

\(\displaystyle \begin{array}{ccccc} & & & & 1 \\ & & -- & -- & -- \\ 2x+1 & ) & 2x \\ & & 2x & + & 1 \\ & & -- & -- & -- \\ & & & - & 1 \end{array}\)


\(\displaystyle \text{Hence: }\;\;\frac{2x}{2x+1} \;\;=\; \;1 - \frac{1}{2x+1}\)