# Integration, u substitution with limits

1. Feb 21, 2012

### sg001

1. The problem statement, all variables and given/known data

Evaluate ∫ 3x /(3x+1)^2.dx , with limits 1 & 0
using the sustitution u = 3x+1

2. Relevant equations

3. The attempt at a solution

u= 3x+1
du/dx = 3
dx = du/3

Therefore,

∫ 3x*(u)^-2 * du/3

= ∫ x* (u)^-2

Since u = 3x +1
Therefore, x = (u-1)/3

Hence,

∫ (u-1)* 1/3*(u^2)

Now, plugging in limits of 1&0 into u

u = 3(1) +1 =4
u = 3(0) + 1 = 1

Therefore, limits of 4 & 1.

Hence,
1/3 ∫ (u^-1)-(u^-2)

=1/3 [-u + 2u^-1] with limits 4& 1.

= 1/3 (- 9/2) = -3/2.

However, there is no answer mathing this solution.
But i dont know where i went wrong??
thankyou.

2. Feb 21, 2012

### ehild

Your integral of $$\int_1^4{(\frac{1}{u}-\frac{1}{u^2})du}$$ is wrong. Look after the basic integral formulae.

ehild

3. Feb 21, 2012

### sg001

Should it be...

ln u + u^-1

4. Feb 21, 2012

### ehild

Yes, it is perfect now!

ehild

5. Feb 21, 2012

### sg001

yay!!

Thanks so much for the help.
Greatly appreciated