Integration

Question 1

what is an improper fraction

Answer 1

numerator ≥ denominator

Question 2

express

6x²+x+1/(x-1)(x+1)² as partial fractions

Answer 2

2/x-1 + 4/x+1 - 3/(x+1)²

Question 3

when splitting into partial fractions

if the top only has an x² and bottom is x

Answer 3

the top is a second degree and bottom is 1st degree

Ax + B + (C/x+c)

Question 4

if top and bottom are both x²

Answer 4

then top and bottom are both second degree

A + (B/(x+c)) + (C/(x+c))

Question 5

express x²+5/(x-1)(x+2) as partial fractions

Answer 5

both are second degree so A + B/(x-1) + C/(x+2)

A(x-1)(x+2) + B(x+2) + C(x-1) =x² +5

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

Question 6

find ∫ x²/(x-2)(x+1) dx

Answer 6

both second degree A + B/(x-2) + C/(x+1)

∫1dx + 4/3∫1/(x-2)dx -1/3∫1/(x+1)dx

x+(4/3)ln|x-2| -(1/3)ln|x+1| +c

Question 7

if numerator ≥ the denominator there will be a

Answer 7

a quotient

Question 8

∫x²-3/(x+2)(x-1)dx

Answer 8

both second degree so A+ B(x+2) + C/(x-1)

∫1dx - (1/3)∫1/(x+2)dx - (2/3)∫1/(x-1)

x-(1/3)ln|x+2|-(2/3)ln|x-1| +c

Question 9

what is the problem with 4—>1∫1(x-2)dx

Answer 9

cannot be evaluated as x=2 gives a disconuity

Question 10

0—>π/2∫cos²2x dx

Answer 10

cosA=cosAcosA-sinAsinA

A=2x—>cos²A-sin²A

cos²A-(-cos²A+1)

cos²A-1

cos4x=2cos²2x-1

cos²2x=(1/2)(cos4x+1)

=(1/2)∫c0s4x+1)

0—->π/8(1/2)[(1/4)sin4x +x]

1/8 + π/16