integration by part

**stevedave** · January 10th 2009, 11:33 AM

I'm having some trouble with this

$\int{\ln{(2x+1)}dx}$

I set $u = ln(2x+1)$ and $dv = dx$ so that $du=\frac{2}{2x+1}$ and $v = x$ .

so after integrating by part once I get:

$x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}$

then I tried to integrate by part again with $u=x$ and $dv=(2x+1)^{-1}$

my final answer ends up as $x\ln{(2x+1)}$ but my solution manual tells me differently.

Can anyone tell me what I'm doing wrong?

**Chris L T521** · January 10th 2009, 11:45 AM

Originally Posted by stevedave

I'm having some trouble with this

$\int{\ln{(2x+1)}dx}$

I set $u = ln(2x+1)$ and $dv = dx$ so that $du=\frac{2}{2x+1}$ and $v = x$ .

so after integrating by part once I get:

$x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}$

then I tried to integrate by part again with $u=x$ and $dv=(2x+1)^{-1}$

my final answer ends up as $x\ln{(2x+1)}$ but my solution manual tells me differently.

Can anyone tell me what I'm doing wrong?

For the integral $-\int\frac{2x}{2x+1}\,dx$ , make the substitution $z=2x+1\implies x=\tfrac{1}{2}\left(z-1\right)$

Thus, $\,dz=2\,dx$ .

The integral becomes: $-\int\frac{x\,dz}{z}\implies -\tfrac{1}{2}\int \frac{z-1}{z}\,dz=\tfrac{1}{2}\int\left(\frac{1}{z}-1\right)\,dz$

Can you continue from here?

**Danny** · January 10th 2009, 11:47 AM

Originally Posted by stevedave

I'm having some trouble with this

$\int{\ln{(2x+1)}dx}$

I set $u = ln(2x+1)$ and $dv = dx$ so that $du=\frac{2}{2x+1}$ and $v = x$ .

so after integrating by part once I get:

$x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}$

then I tried to integrate by part again with $u=x$ and $dv=(2x+1)^{-1}$

my final answer ends up as $x\ln{(2x+1)}$ but my solution manual tells me differently.

Can anyone tell me what I'm doing wrong?

In the second integral, expand

$\int\frac{x}{2x+1}dx = \int \frac{1}{2} - \frac{1}{2}\frac{1}{2x+1} dx$ and integrate each term separately

BTW $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln| ax + b| + c$

**skeeter** · January 10th 2009, 11:51 AM

$\int \ln(2x+1) \, dx = x\ln(2x+1) - \int \frac{2x}{2x+1} \, dx<br />$

note that $\frac{2x}{2x+1} = \frac{2x+1-1}{2x+1} = 1 - \frac{1}{2x+1}$

$\int \ln(2x+1) \, dx = x\ln(2x+1) - \int 1 - \frac{1}{2x+1} \, dx$

can you finish from here?

**stevedave** · January 10th 2009, 11:56 AM

Thank you everyone. That was very helpful.

**tom@ballooncalculus** · January 10th 2009, 12:21 PM

How about this for a short cut? Notice that one integral of 1 is x + 1/2, and make that the value of v, for dv/dx = 1 ...

As usual, straight lines differentiate down / integrate up with respect to x.

Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers

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