integral ((2^2-4t-5)/(t-1))dt?

Answer 1

(I assume "2^2" is a typo for 2t^2)

With polynomials, unless you see a common factor or a substitution that simplifies, you just divide in, get each term and integrate it separately.

I = Integ ((2t²-4t-5)/(t-1))dt

(2t²-4t-5) doesn't have factors with integer coefficients, so there is no common factor with (t-1). We could certainly do the general partial fraction or synthetic division approaches, and they would both work, but this one is screaming out for the simplifying substitution:
set u = t-1
du = dt

I = Integ ((2(u+1)²-4(u+1)-5)/u)du
I = Integ ((2u² + 4u +1 - 4u -4 - 5)/u) du ; [Ok Fred spotted a typo here]
= Integ ((2u² -8)/u) du
= Integ (2u - 8/u) du
= u² - 8 ln (u) + C

Thus, substituting back:
I = Integ ((2t²-4t-5)/(t-1))dt
= (t-1)² - 8 ln (t-1) + C


[Ok Fred spotted my typo but I got the correct method first, and the correct answer off by 1]

Answer 2

There is a small mistake in the second line of the substitution of the previous answer and the 1 has not been doubled. The 8 should be a 7.

By far the simplest way to deal with this question is simple long division of 2t² - 4t - 5 by t - 1

(2t² - 4t - 5)/(t - 1) = 2t - 2 - 7/(t - 1)

∫ (2t² - 4t - 5)/(t - 1) dt = ∫ 2tdt - ∫ 2dt - ∫ 7/(t - 1) dt

.................................... = t² - 2t - 7ln(t - 1) + c

Answer 3

Assume question is:-
I = ∫ (2t² - 4t - 5) / (t - 1) dt
(2t² - 4t - 5) / (t - 1) = (2t - 2) - 7 / (t - 1) upon division being carried out.
I = ∫ (2t - 2).dt - 7 ∫1 / (t - 1) dt
I = t² - 2t - 7 log (t - 1) + C

Answer 4

i believe this is the way how to solve it:

you can simplify the problem by actually dividing the numerator with the denominator:

(2t^2-4t -5)/(t-1)

2t - 2 -7/(t-1)
. . . . .-----------------
(t-1) / 2t^2 - 4t - 5
. . . . . 2t^2 +2t
. . . . --------------
. . . . . . . . . . . -2t - 5
. . . . . . . . . . . .2t - 2
. . . . . . . . . . . .---------
. . . . . . . . . . . . . . .-7

the result is [ 2t - 2 - 7/(t-1) ]dt

integrating each term will be

integral{ 2tdt - 2dt - [7/(t-1)]dt}

answer would be:

t^2 - 2t - 7 ln (t-1) + C