# Evaluate the definite integral of f(x)=x^2/(x^2 - 6x + 10) from x=3/2 to x=2.

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To calculate the definite integral of the given function, we'll use the substitution method.

We'll calculate Integral of f(x) = x^2/(x^2 - 6x + 10)

We notice that if we'll complete the numrator, we'll obtain the denominator, x^2 - 6x + 10, so, we'll subtract and add 6x and we'll add and subtract 10.

The numerator will become:

x^2 - 6x + 10 + 6x - 10

The integral will become:

Int (x^2- 6x + 10 + 6x - 10)dx/(x^2 - 6x + 10)

We'll re-write the integral, using the addition property of integrals:

Int (x^2- 6x + 10)dx/(x^2 - 6x + 10)+Int (6x - 10)dx/(x^2 - 6x + 10)

We'll solve the first integral:

Int (x^2- 6x + 10)dx/(x^2 - 6x + 10) = Int dx = x + C

We'll solve the second integral:

Int (6x - 10)dx/(x^2 - 6x + 10)

We'll create the square (2x-3)^2 to the denominator.

x^2 - 6x + 10 = (2x-3)^2 + 31

We'll note 2x-3= t.

We'll write the numerator in t:

2x-3= t

2x = t+3

x = (t+3)/2

6(t+3)/2 - 10 = 3(t+3) - 10

We'll remove the brackets:

3t+9-10

The numerator will become:

6x - 10 = 3t-1

We'll re-write the integral in t:

Int (6x - 10)dx/[(2x-3)^2 + 31] = Int (3t-1)dt/(t^2 + 31)

Int (3t-1)dt/(t^2 + 31) = 3Int dt/(t^2 + 31) - Int dt/(t^2 + 31)

3Int dt/(t^2 + 31) = 6/2 ln (t^2 + 31) + C

Int dt/(t^2 + 31) = (2/sqrt31)*arctan (1/sqrt31)

**Int f(x)dx = 1/3 + 3ln (32/31) - (2/sqrt31)*arctan (1/sqrt31)**

To evaluate the definite integral f(x) = x^2/(x^2-6x+10) from x =3/2 to x=2.

We know that x^2/(x^2-6x+10) = 1 qotient and remainder 6x-10.

6x-10 = 3(2x-6) + 8.

Therefore x^2/(x^2-6x+10) = 1 + [3(2x-3)+8]/(x^2-6x+10)

Therefore Int x^2dx/(x^2 -6x+10) = int {1 }dx+ Int3(2x-6)dx/(x^2-6x+10) + Int {8/(x^2-6x+10}dx = I1 +I2+I3

I1 = x. Taking the limits from x= 3/2 to 2, we get: I1(2)-I1(3/2) = 2-3/2 = 1/2.

I2 = 2dt/t, where t = x^2-6x+10, dt = (x^2-6x+10)' = (2x-6)dx, when x = 0, t = (3/2)^2 - 6(3/2)+10 = 9/4-9+10 =13/4. when x=2, t = 2^2-6*2+10 = 2.

I2 = 2dt/t = 2lnt . Taking limits I2(2) - I2(2)-I2(13/4) = 2*{ln (2) -ln(13)/4} = 2 ln { 2/ (13/4)} = 2ln(8/13).

I3 = 8dx/(x^2-6x+10) = 8dx/{(x-3)^2+1} = (8/1 ) arctan[ (x-3)/1] = 28arc tan {(x-3)}

I3 (2) - I3 (3/2) = 8 [arctan (4-3)- arctan(3/2-3)] = 2{arctan(-1)+arctan(1/2)} = -8{arctan (1) - arctan(1/2) }

Therefore, I1+T2+I3 = Int f(x) dx = (1/2) +2ln(8/13) - 2{ pi/4 - arctan(1/2)}