Integrating by Substitution: Evaluating \int \frac{3x}{x^2+2}

[Radarithm]

Homework Statement

Evaluate: \int \frac{3x}{x^2+2}

Homework Equations

\int \frac{1}{u} \frac{du}{dx} dx= \ln u + C

The Attempt at a Solution

I got a horribly wrong answer: \frac{1}{2x}\ln (x^2+2)+C
This was done by multiplying \frac{du}{dx} by \frac{3x}{u}
This part is what confuses me: When the book shows an example, they multiply the integral by 1 over whatever number they multiplied the numerator with; for example:
\int \frac{x}{x^2+1} dx = \frac{1}{2} \int \frac{2x}{x^2+1} dx = \frac{1}{2} \int \frac{1}{u} \frac{du}{dx} dx
= \frac{1}{2} \ln u + C = \frac{1}{2} \ln (x^2+1) +C
The correct answer given by the book for my problem seems to be \frac{3}{2} \ln (x^2+2) + C
I need help with integrating by substitution. I still fail to see how the above example from the book makes sense. Doesn't the chain rule say that you must multiply du by \frac{du}{dx}? Are they somehow trying to cancel something out? I fail to see what exactly they're doing.

 

[vanhees71]

Try to substitute
u=x^2+2.
Note that this implies
\mathrm{d} u = \mathrm{d} x \; 2 x.

 

[Radarithm]

Try to substitute
u=x^2+2.
Note that this implies
\matrm{d} u = \mathrm{d} x 2 x.

I did, but I don't know what to do after getting here:

\int \frac{3x}{u} dx
A when I multiply by \frac{du}{dx} I get: \int \frac{5x^2}{u} dx
I do not know how the book got \frac{3}{5} \ln (x^2+2)+C

edit: So am I supposed to "cancel out" the 2x? That is what I think they did in the example, except it was 2du

 

[Office_Shredder]

You did the du/dx substitution backwards. Once you have
\int \frac{3x}{u} dx

You can observe that
\frac{du}{dx} = 2x
to get after some basic algebra
x dx = \frac{1}{2} du

Now you just need to replace the xdx in your integral by 1/2 du and you have an integral you should be able to solve.

 

[Radarithm]

Now you just need to replace the xdx in your integral by 1/2 du and you have an integral you should be able to solve.

I don't understand what you mean by that; should I replace the 3xdx with \frac{du}{2}?
Sorry if I'm being annoying, I'm just new to integrals.

 

[Ray Vickson]

I don't understand what you mean by that; should I replace the 3xdx with \frac{du}{2}?
Sorry if I'm being annoying, I'm just new to integrals.

No, that is not what he said/wrote. He said $x dx = \frac{1}{2} du$. How would you re-write $3 x dx$? (Don't guess: sit down and work things out carefully, step-by-step.)

 

[Radarithm]

No, that is not what he said/wrote. He said $x dx = \frac{1}{2} du$. How would you re-write $3 x dx$? (Don't guess: sit down and work things out carefully, step-by-step.)

I think I finally got it:
xdx=\frac {1}{2}du3xdx=\frac {3}{2}du
\frac {3}{2}\int u \frac{du}{dx}dx= \frac{3}{2} \ln u= \frac{3}{2} \ln (x^2+2)+C
So I need to find xdx and multiply the integral by nxdx to get the anti-derivative?