# Integrating (5x)/(3x^2+5) dx

1. Nov 2, 2012

### beaf123

Hi all. I am having Calculus 1 this year. We are using a book called Thomas Calculus.
I think its a lot of fun, but I have to work very much since there is basic stuff like trigonometry that I know really bad. Since I work so much with math I thought it could be fun and helpful to talk with other math people in here:-)

To the question:

∫ (5x)/(3x^2+5) dx

∫ (1/(3x^2+5)) * 5x

Integration by parts give.

(5x) ln(3x^2+5) - 5 ∫ ln (3x^2+5)dx

Not sure have to calculate the last integral. Not sure about anything here really..

2. Nov 2, 2012

### hedipaldi

compute by substitution:u=3x^2+5

3. Nov 3, 2012

A new question: With u=3x^2+5 the answer is 5/6*log(3*x^2+5), buit using symbolics in matlab the answer is 5/6*log(x^2+5/3) and that's using u=x^2+5/3. Why are there different answers?
As 5/6*log(3*x^2+5) = 5/6*log(x^2+5/3)/log(3) <> 5/6*log(x^2+5/3)

4. Nov 3, 2012

### SammyS

Staff Emeritus
The two answers differ only by a constant. Remember the constant of integration?

$\displaystyle \frac{5}{6}\log(3x^2+5)=\frac{5}{6}\log\left(3 \left(x^2+\frac{5}{3}\right)\right)$
$\displaystyle =\frac{5}{6}\log(3)+\frac{5}{6}\log\left(x^2+\frac{5}{3}\right)$​

5. Nov 3, 2012

### beaf123

Yes of course. I should have thought of that.
Would you get the same answer using integration by parts?

If the exercise looked like this instead:

∫ (5x)/(3x^2+4x+5) dx

then you have to use integration by parts?

6. Nov 3, 2012

### tiny-tim

welcome to pf!

hi beaf123! welcome to pf!

(try using the X2 button just above the Reply box )
integrating by parts, your u would be the whole thing, and your v would be 1

(your line starting "(5x) ln(3x^2+5) …" was wrong)
no, write the integrand A(6x+4)/(3x2+4x+5) + B/(3x2+4x+5), and do two different substitutions

7. Nov 6, 2012

Thank you:)