evaluate the integral?

Question

please show steps

integrate upper limit 1 lower limit 0 for (x-1)/(x^2+3x+2) dx

Answer 1

First you factor the quadratic

x² + 3x + 2 = (x+2)(x+1)

Using partial fractions we can say that

(x-1)/(x² + 3x + 2) = A/(x+2) + B/(x+1)

(x-1)/(x² + 3x + 2) = A(x+1)/(x+2)(x+1) + B(x+2)/(x+1)(x+2)

from this we get

x-1 = A(x+1) + B(x+2)

x-1 = (A+B)x + (A+2B)

We need to equate the coefficients for x and the constant coefficient.

so A+B=1 and A+2B = -1

substract A+B from A+2B

A+2B - (A+B) = -1 -1

B = -2 and A=3

so (x-1)/(x² + 3x + 2) = 3/(x+2) - 2/(x+1)

to check 3(x+1) - 2(x+2) = x-1 as required.

Now integrate

∫(x-1)/(x² + 3x + 2)dx = 3∫dx/(x+2) - 2∫dx/(x+1)

the integrals on the right are of the form du/u and ∫du/u = ln(u)

3∫dx/(x+2) - 2∫dx/(x+1) = 3ln(x+2) - 2ln(x+1)|0,1

= 3ln(1+2) - 2ln(1+1) - 3ln(0+2) + 2ln(0+1)
= 3ln(3) - 2ln(2) - 3ln(2) + 2ln(1)
= 3ln(3) - 5ln(2) + 2ln(1)

ln(1)=0 and aln(b) = ln(b^a)

= ln(3^3) - ln(2^5)
= ln(3^3/2^5)
= ln(27/32)

so using a calculator you can find ln(27/32) = -0.17

Answer 2

x^2 + 3x + 2 = (x+1)(x+2)

A/(x+1) + B/(x+2) = (x-1) /(x+1)(x+2)

Find A and B and then integrate the simple fractions

Ana

Answer 3

(x - 1) / (x + 2).(x + 1) = A/(x + 2) + B/(x + 1)
(x - 1) = A.(x + 1) + B.(x + 2)
1 = A + B
-1 = A + 2B

1 = A + B
1 = - A - 2B----ADD
2 = - B
B = - 2
A = 3

I = 3 ∫ 1 / (x + 2) dx - 2 ∫ 1 / (x + 1) . dx
I = 3.log (x + 2) - 2.log (x + 1)
Limits are 0 to 1:-
I = 3.[log (x + 2)] - 2.[log(x + 1)] lims 0 to 1
I = 3.(log 3 - log2) - 2(log 2 - log1)
I = 3.log 3 - 5.log 2 + 2 log 1
I = 3.log 3 - 5.log 2

Answer 4

Express (x-1)/[(x+1)(x+2)] in partial fractions
= A/(x+1) + B/(x+2)
= [ A(x+2) + B(x+1) ] / [(x+1)(x+2)]

So A+B = 1
2A + B = -1
A = -2, B = 3

Now we need to integrate -2/(x+1) + 3/(x+2)
Integral = -2*ln(x+1) + 3*ln(x+2)

Apply limits to get
[-2*ln2 + 3*ln3] - [-2*ln1 + 3*ln2]
= 3*ln3 - 5*ln2
= -0.1699