Log integration question (1 Viewer)

[Dragie]

Hi, I have no idea how to do these questions:

a) e
∫ 2t + (2/t) dt
1



b) Show that (x+7)/(x-1) can be written as 1 + (8/x-1) and show that
e+1
∫ (x + 7)/(x - 1) = e + 7
2

Answers:
a)e² + 1

 

[Riviet]

For a), bring the 2 out of the integral, and remember that ∫1/t dt = log_et, no constant needed since it's a definite integral.

b) (x+7)/(x-1)=(x-1+8)/(x-1)
= (x-1)/x-1) + 8/(x-1)
= 1 + 8/(x-1)

So when integrating (x+7)/(x-1), change this to what we've just changed the expression into, 1 + 8/(x-1), and integrate that.

Remember that ∫a.f '(x)/f(x) dx = a.log_e[f(x)], no constant needed for same reason as before. I hope that helps.

 

[Trev]

a)
[t²+2ln(t)] {1->e}
=e²+2ln(e)-1²-2ln(1)
=e²+2-1-0
=e²-1

b)
(x+7)/(x-1) = (x-1+8)/(x-1) =1+8/(x-1), or you can do this by division.
Then, integrating:
int&; [1+8/(x-1)].dx {2->e+1}
= [x+8ln(x-1)] {2->e+1}
= e+1+8ln(e+1-1)-2-8ln(2-1)
= e+1+8-2-0
= e+7, as required.

 

[Dragie]

Oooh thank you!

I just have one more question:

(x²e^2x2 - 1)/x

 

[Riviet]

Split the fraction up into xe^2x2 and -1/x and use separate integration signs for each bit. Integrating the 1/x is easy, but for the other bit, use the substitution u=x², du=2x dx, and see if you can finish it off.

 

[Dragie]

Oh awesome , you've helped me heaps thank you!