Integration, u substitution with limits

[sg001]

1. Homework Statement

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

Homework Equations

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 don't know where i went wrong??
please help.
thankyou.

 

[ehild]

...
Therefore, limits of 4 & 1.

Hence,
1/3 ∫ (u^-1)-(u^-2)

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

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


ehild

 

[sg001]

Should it be...

ln u + u^-1

 

[ehild]

Should it be...

ln u + u^-1

Yes, it is perfect now!

ehild

 

[sg001]

yay!

Thanks so much for the help.
Greatly appreciated