# Basic Integral

1. Jan 21, 2014

1. The problem statement, all variables and given/known data
Evaluate: $$\int \frac{3x}{x^2+2}$$

2. Relevant equations
$$\int \frac{1}{u} \frac{du}{dx} dx$$$$= \ln u + C$$

3. 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.

Last edited: Jan 21, 2014
2. Jan 21, 2014

### vanhees71

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

3. Jan 21, 2014

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 $2du$

4. Jan 21, 2014

### Office_Shredder

Staff Emeritus
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.

5. Jan 21, 2014

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.

6. Jan 21, 2014

### Ray Vickson

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.)

7. Jan 21, 2014

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