Maths Integration Problem

adam_ccfc

Any help appreciated!

It's from the 1997 Higher Level Exam.

Paper 1, Question 8, Part (c):

Calculate the value of


I know it involves substitution and letting u = √t, but it's not working out for me!

PeterMC

Take the Root T outside the bracket so its root t(root t + 1). Substitute in then ... I think ;-)

adam_ccfc

Sound man, I'll give it a go!

breadmonkey

I don't ever remember doing an integral like this.

ApeXaviour

breadmonkey wrote:

I don't ever remember doing an integral like this.

I remember it. That is to say I remember not being able to do it. 4 years of physics later and you'd think I'd have an easier time? Well.. I did but it's still tricky if you're out of practice.

OP did you get it out? The answer I got is attached. I substituted for root(t) + 1. Though I'm 7 months out of practice with calculus so there's a high possibility I made a blunder somewhere

Edit: Just noticed OP is banned. Oh well...

breadmonkey

I got almost that answer except I had 1/2(ln_) as opposed to 2ln_

RefulgentGnomon

Its 2ln[(sqrt3 + 1)/(1/sqrt3 + 1 )

which

= 2ln [(sqrt3 +1)]/[(sqrt3 + 1)/sqrt3]

= 2ln [sqrt3 + 1]*[(sqrt3)/(sqrt3 + 1)]

(sqrt3 + 1) terms cancel


2ln(sqrt3)

= ln (sqrt3)^2

=ln3


! (was going to put that exclamation mark up further but maybe it would have looked like ln(3 factorial)...;) )

JC 2K3

int(1/(t+√t) = int(1/(√t(√t+1)

u = √t

Getting the integration constants for u:
u=√(3) and √(1/3)


du = dt/2√t
2du = dt/√t

(2)int(1/(u+1))=[2(ln(u+1))]

[2(ln(√3+1) - ln(1/√3 + 1))]
=2(ln((√3+1)/(1/√3 + 1)))
=2(ln((√3+1)/((1+√3)/√3)))
=2(ln(1/(1/√3)))
=2(ln(√3))
=ln(√3)^2
=ln3


Yeah, so everyone was right on this thread.