I'm not sure how to approach this...

**mortalapeman** · July 20th 2008, 10:31 PM

My problem is I don't i don't know what to substitute in for f(x) and g(x) when i try to use integration by parts for this particular integral.

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

I can kind of get thorough it about halfway if i use $f(x)=ln(2x+1)$ and $g(x)=dx$ . but i end up with my second integral being this:

$\int \frac{2x}{2x+1}dx$

and i'm not sure how to get

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

As my answer if i integrate that last one. I would post all my work but i'm kind of in a hurry. So thank you and I appreciate your help.

**CaptainBlack** · July 20th 2008, 10:44 PM

Originally Posted by mortalapeman

My problem is I don't i don't know what to substitute in for f(x) and g(x) when i try to use integration by parts for this particular integral.

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

I can kind of get thorough it about halfway if i use $f(x)=ln(2x+1)$ and $g(x)=dx$ . but i end up with my second integral being this:

$\int \frac{2x}{2x+1}dx$

and i'm not sure how to get

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

As my answer if i integrate that last one. I would post all my work but i'm kind of in a hurry. So thank you and I appreciate your help.

$\int \frac{2x}{2x+1}dx=\int 1-\frac{1}{2x+1}~dx$

RonL

**mortalapeman** · July 20th 2008, 10:55 PM

Originally Posted by CaptainBlack

$\int \frac{2x}{2x+1}dx=\int 1-\frac{1}{2x+1}~dx$

RonL

What exactly do you mean? Are you saying that those two expressions are equal or that i need to substitute yours in for mine because i messed up somewhere along the way and that is the integral i should have got?

Also, did you use the same substitutions as me?

**Chris L T521** · July 20th 2008, 10:59 PM

Originally Posted by mortalapeman

My problem is I don't i don't know what to substitute in for f(x) and g(x) when i try to use integration by parts for this particular integral.

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

I can kind of get thorough it about halfway if i use $f(x)=ln(2x+1)$ and $g(x)=dx$ . but i end up with my second integral being this:

$\int \frac{2x}{2x+1}dx$

and i'm not sure how to get

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

As my answer if i integrate that last one. I would post all my work but i'm kind of in a hurry. So thank you and I appreciate your help.

$\int\ln(2x+1)\,dx$

Let $u=\ln(2x+1)$ and $\,dv=\,dx$ . Thus $\,du=\frac{2\,dx}{2x+1}$ and $v=x$ .

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

To integrate $\int\frac{2x\,dx}{2x+1}$ , let $u=2x+1\implies x=\frac{u-1}{2}$

Thus, $\,du=2\,dx$

The integral is transformed into $\frac{1}{2}\int\frac{u-1}{u}\,du=\frac{1}{2}\int\left[1-\frac{1}{u}\right]\,du=\frac{1}{2}u-\frac{1}{2}\ln|u|+C$

Resubstitute:

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

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

$\implies \left(x+\frac{1}{2}\right)\ln(2x+1)-\frac{1}{2}(2x+1)+C$

$\implies \frac{1}{2}(2x+1)\ln(2x+1)-x\underbrace{-\frac{1}{2}+C}_{C}$

$\implies \color{red}\boxed{\frac{1}{2}(2x+1)\ln(2x+1)-x+C}$

The $\frac{1}{2}$ and $C$ make up another constant, which I called $C$ ...

I hope that this clarifies things!

--Chris

**mortalapeman** · July 20th 2008, 11:03 PM

Originally Posted by Chris L T521

$\int\ln(2x+1)\,dx$

Let $u=\ln(2x+1)$ and $\,dv=\,dx$ . Thus $\,du=\frac{2\,dx}{2x+1}$ and $v=x$ .

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

To integrate $\int\frac{2x\,dx}{2x+1}$ , let $u=2x+1\implies x=\frac{u-1}{2}$

Thus, $\,du=2\,dx$

The integral is transformed into $\frac{1}{2}\int\frac{u-1}{u}\,du=\frac{1}{2}\int\left[1-\frac{1}{u}\right]\,du=\frac{1}{2}u-\frac{1}{2}\ln|u|+C$

Resubstitute:

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

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

$\implies \left(x+\frac{1}{2}\right)\ln(2x+1)-\frac{1}{2}(2x+1)+C$

$\implies \frac{1}{2}(2x+1)\ln(2x+1)-x\underbrace{-\frac{1}{2}+C}_{C}$

$\implies \color{red}\boxed{\frac{1}{2}(2x+1)\ln(2x+1)-x+C}$

The $\frac{1}{2}$ and $C$ make up another constant, which I called $C$ ...

I hope that this clarifies things!

--Chris

Wow thank you so much, you have no idea >.< now i can work those other 10 problems just like that.

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