Question

Integrate
(3x^3-x^2+2x-4)/(x^2-3x+2)^0.5 from 0 to 1......( only the denominator is in square root)

Answer 1

Q: Integrate from 0 to 1 :
$\frac{3 {x}^{3} - {x}^{2} + 2x - 4}{ \sqrt{ {x}^{2} - 3x + 2} }$
Final Answer :
$\frac{1}{16} ( - 202 \sqrt{2} - 135 ln(3 - 2 \sqrt{2} )$
Solution :
Steps and Understanding :

1) Since, we observe that degree of expression in Numerator is higher than that of Denominator.
So, We will divide by denominator without root with Numerator to find quotient and remainder.
Then, we got.
$(3 {x}^{3} - {x}^{2} + 2x - 4) = ( {x}^{2} - 3x + 2) (3x + 8) \\ + 20(x - 1)$

2) Substitute these parts in original definite integral.
Write required parts obtained in different integrals.
as shown in pic.

I(1) = First part.
I(2) = Second part.
We will try to find the integrals separately.


3) These definite integrals obtained by converting them into standard Integrals.
First Part :
Use :

$(3x + 8) =a \ \times \frac{d}{dx} ( {x}^{2} - 3x + 2) \\ + b$
where A, B are constant.
After solving,
A = 3/2 , B = 25/2


4) After simplification of first part ,then
we will get their Integrals as shown in pic which are standard integrals.


After solving we will get,
I(1) =
$\frac{1}{16} (118 \sqrt{2} + 25 ln(3 - 2 \sqrt{2} ) )$

5) Now comes second part, I(2)

Now,
$(x - 1) = c \times \frac{d}{dx} ( {x}^{2} - 3x + 2) + d \:$
Here after solving as shown in pic, we will get
c = 1/2 , d = 1/2


6) Now, After substitution Second integral goes simplified as they are converted into standard integrals.

After solving
I(2) =
$- \sqrt{2} - \frac{1}{2} ln(3 - 2 \sqrt{2} )$



7) Required
I = I(1) + 20I(2)
After substitution.
$= \frac{1}{16} ( - 202 \sqrt{2} - 135 ln(3 - 2 \sqrt{2} ) )$