Integration with fraction like equations...

Question

Integrate ∫ (10x^2 -45x +6) / (2x^2 - 9x + 4 ) dx

Answer 1

First use polynomial division to divide the two polynomials:

_____________

2x²- 9x + 4 | 10x² - 45x + 6 = 5 with remainder -14

or, in other words 5 - 14/(2x²- 9x + 4)

Now use partial fraction decomposition to break 14/(2x²- 9x + 4) into two fractions. First factor the denominator:

14/(2x²- 9x + 4) = 14/[(2x - 1)(x - 4)]

14/[(2x - 1)(x - 4)] = A/(2x - 1) + B/(x - 4)

A(x-4)/[(2x - 1)(x - 4)] + B(2x - 1)/[(2x - 1)(x - 4)] must equal 14/[(2x - 1)(x - 4)]

So A(x - 4) + B(2x - 1) = 14

Ax + 2Bx = 0 and -4A - B = 14

Solve to get A = -4 and B = 2

Therefore 14/[(2x - 1)(x - 4)] = -4/(2x - 1) + 2/(x - 4)

So the new equivalent integral is:

∫5dx - ∫[-4/(2x - 1) + 2/(x - 4)] dx = ∫5dx + ∫4/(2x - 1) dx - ∫2/(x - 4) dx

Take the antiderivative to get:

5x + 2ln(2x-1) - 2ln(x - 4) + C

Answer 2

I = ∫[5(2x² - 9x + 4) - 14]/(2x² - 9x + 4)dx = ∫[5 - 7/(x² - 9/2x + 2]dx = 5x - ∫7/[(x - 9/4)² - 49/16]dx = 5x -

7/(7/2)lnI(x - 9/4 - 7/4)/(x - 9/4 + 7/4)I + C = 5x - 2lnI(x - 4)/(x -1/2)I + C

Answer 3

I would do this as follows:

Divide the numerator by the denominator by long division to get

5-14/(2x²-9x+4)≡5+(A/(x-4))+(B/(2x-1))

A and B are determined by partial fractions as -2 and 4 respectively

Then integrate as 5x and 2 terms in the natural logarithm.