Indefinite Integrals

Question

My daughter is having a hard time understanding this and it's above my head. Any help to show her/me how to work this is greatly appreciated.

Consider the indefinite integral ((6x^3+3x^2-1x-6)/(x^2-1))dx.

Then the integrand decomposes into the form

ax+b+(c/x-1)+(d/x+1)

where

a=?

b=?

c=?

d=?

Integrating term by term, we obtain that

integral ((6x^3+3x^2-1x-6)/(x^2-1))dx =?

We need to solve/answer everywhere there is a "?". Thanks in advance.

Answer 1

Using long division we can get

(6x³ + 3x² – x – 6)/(x² – 1) = 6x + 3 + (5x – 3)/(x² – 1)

(So a = 6 and b = 3)

We finish by expressing the last term as partial fractions

(5x – 3)/(x² – 1) = (5x – 3)/[(x – 1)(x + 1)]

So we write down

(5x – 3)/[(x – 1)(x + 1)] = c/((x – 1) + d/(x + 1) and solve for c and d

Multiplying both sides by (x – 1)(x + 1) gives

5x – 3 = c(x + 1) + d(x – 1) = (c + d)x + (c – d)

This gives us the system

c + d = 5

c – d = –3

Adding the two equations eliminates d and gives

2c = 2 which means that c = 1

Since c + d = 5 we can write 1 + d = 5 which means that d = 4

So (finally!) our integrand decomposes to

(6x³ + 3x² – x – 6)/(x² – 1) = 6x + 3 + 1/(x – 1) + 4/(x + 1)

a = 6

b = 3

c = 1

d = 4

Integrating term by term (omitting the constant of integration until the end)

∫6xdx = 6(x²/2) = 3x²

∫3dx = 3x

∫dx/(x – 1) = ln|x – 1|

∫4dx/(x + 1) = 4ln|x + 1|

Answer

∫[(6x³ + 3x² – x – 6)/(x² – 1)]dx = 3x² + 3x + ln|x – 1| + 4ln|x + 1| + C