C3 Intergration problem

How far have you got with an attempt?

Take the derivative of U, which should let you replace $dx$ with $du$. Which might just let you cancel out a term or two..

u=2x+1 => du/dx = 2. Therefore we want to integrate (u-1)/2 * 1/u dx(substituting in for x) or (u-1)/2 * 1/u *1/2 du = (u-1)/4u = 1/4 - 1/4u, integrating goes to u/4 + ln(4u)/4. = (2x+1)/4 - ln(8x+4)/4. Substituting in x=1 and x=0 gives a value of:

3/4-ln(12)/4 -1/4 +ln(4)/4 = 1/4[3-1+ln(4)-ln(12)] = [2+ln(4/12)]/4 = [2+ln(1/3]/4, bear in mind that ln(1/3)=ln(3^-1)=-ln(3) andyou'll get you're answer

I have let u = 2x+1 so sube the bottom half of the equation with U, I then rearranged this to get X for the top

U = 2X + 1 so X= (u-1)/2 sub thi at the top

du/dx =2 so du = 2dx therefore i take a half outside the integral, this leaves me with

(integral sign here) U -1 du
U

After this I changed the limits using u=2x+1 to get the new limits u=3 and u=1 this is where i got stuck

Yep that's right, just seperate it into $1 - \frac{1}{u}$ and integrate with respect to u. Once you've done that substite 2x + 1 back in for u. This way you don't have to change your limits of integration. Don't forget the 1/4 that you'll have taken outside the integral.