Question

Solution to the following integral (in post below)

Answer 1

\[\int\limits_{}^{} \frac{t-1}{t^{2}+4t+3} dt\]
I'm kind of stuck on this one, i've tried a few different methods of u substitution but can't solve it.

Answer 2

Does partial fraction help?

Answer 3

Yeah i think that might be the answer. I have only done partial fractions in the form where there is a coefficient and a variable however. Not when there are two terms in the numerator.

Answer 4

Hmmm...
\[\frac{t-1}{t^{2}+4t+3}=\frac{t-1}{(t+1)(t+3)} = \frac{A}{t+1}+\frac{B}{t+3}\]
So,
A(t+3) + B(t+1) = t-1
Comparing the coefficients,
A+B = 1
3A +B = -1

Something like this?!

Answer 5

Yep... i think i jumped the gun when i gave up on trying to solve with partial fractions. I just revised my notes and i'll have a go using partial fractions now. I'll let you know how i go.

Answer 6

Okay!!~

Answer 7

Awesome, partial fractions was the key. Thanks for that =)

Answer 8

You're welcome :)

Answer 9

Use partial fraction method..

Answer 10

or you may use partial fractions alternatively, what i mean is
t-1 / t^2 +4t +3
= 1/2 ( 2t - 2)/(t^2 +4t +3)
=1/2 ( 2t +4 -6)/(t^2 +4t +3)
= 1/2 ( 2t +4)/(t^2 +4t +3) - 1/2 (6/(t^2 +4t +3))

integrate separately, for 1st part, if you let denominator =z,. you directly get numerator = dz ..

we are concerned with integrating 1/(t^2 +4t +3)
that's also simple, we see we have

1/(t+1)(t+3) = 1/2 (2 / (t+1)(t+3))
=1/2 ( t+3 - (t+1))/(t+1)(t+3))

=>now when you separate out, you'll directly be able to integrate ..hope that helps..

note that i've used partial fractions only,but in a different manner..