Question

int (x^2 - 3x + 7)/(x^2 - 4x + 6)^2 dx      Evaluate the integral

Accepted Answer

You may rebuild the structure you find at denominator, such that:
int (x^2-3x+7)/((x^2-4x+6)^2)dx = int ((x^2-4x+6) + x + 1)/((x^2-4x+6)^2)dx
Separate into two integrals:
int (x^2-4x+6)/((x^2-4x+6)^2)dx + int (x + 1)/((x^2-4x+6)^2)dx
Take the first integral and reduce like terms:
int (x^2-4x+6)/((x^2-4x+6)^2)dx = int 1/((x^2-4x+6))dx  
You may write x^2 - 4x + 6 = x^2 - 4x + 4 + 2 = (x- 2)^2 + 2
int 1/((x^2-4x+6))dx= int 1/((x- 2)^2 + 2)dx = sqrt2/2 arctan ((x - 2)/sqrt2) + c 
You need to take the integral int (x + 1)/((x^2-4x+6)^2)dx  and to separate it into two simpler integrals:
int (x + 1)/((x^2-4x+6)^2)dx = int x/((x^2-4x+6)^2)dx + int 1/((x^2-4x+6)^2)dx 
You should notice that if you differentiate x^2-4x+6 yields 2x - 4, hence, you need to multiply and divide by two and then subtract and add 4, such that:
(1/2)int ((2x-4) + 4)/((x^2-4x+6)^2)dx = (1/2)int ((2x-4))//((x^2-4x+6)^2)dx + 2int 1/((x^2-4x+6)^2)dx 
Hence, int (x + 1)/((x^2-4x+6)^2)dx = (1/2)int ((2x-4))//((x^2-4x+6)^2)dx + 3int 1/((x^2-4x+6)^2)dx
You need to use substitution to solve (1/2)int ((2x-4))//((x^2-4x+6)^2) dx such that:
x^2-4x+6 = t => (2x-4)dx = dt 
(1/2)int ((2x-4))//((x^2-4x+6)^2) dx= (1/2) int (dt)/t^2 = -1/(2t) = -1/(2(x^2-4x+6)) + c 
Put (x - 2)/(2sqrt2)=t => (dx)/(2sqrt2) = dt
3int 1/((x^2-4x+6)^2)dx = 3(sqrt2/4)int 1/((t^2+1)^2) 
Put t = tan alpha => dt = 1/(cos^2 alpha) d alpha
3(sqrt2/4)int 1/((t^2+1)^2) = 3(sqrt2/4)int cos^2 alpha d alpha 
Use the half angle identity:
cos^2 alpha = (1 + cos 2 alpha)/2
3(sqrt2/4)int cos^2 alpha d alpha = 3(sqrt2/4)int(1 + cos 2 alpha)/2 d alpha 
3(sqrt2/8)int d alpha + 3(sqrt2/8) int cos 2 alpha d alpha 
3(sqrt2/8)int d alpha + 3(sqrt2/16) sin 2 alpha 
alpha = arctan((x-2)/(sqrt2))
3(sqrt2/8)arctan((x-2)/(sqrt2)) + 3(sqrt2/16) sin 2 (arctan((x-2)/(sqrt2))) + c 
Hence, evaluating the integral yields int (x^2-3x+7)/((x^2-4x+6)^2)dx =sqrt2/2 arctan ((x - 2)/sqrt2) -1/(2(x^2-4x+6)) +3(sqrt2/8)arctan((x-2)/(sqrt2)) + 3(sqrt2/16) sin 2 (arctan((x-2)/(sqrt2))) + c

Math

`int (x^2 - 3x + 7)/(x^2 - 4x + 6)^2 dx` Evaluate the integral

Expert Answers

Unlock This Answer Now

Related Questions

Evaluate the integral integrate of ((x^3-4x^2+3x+1)/(x^2+4))dx

`int x^2 /(3 + 4x - 4x^2)^(3/2) dx` Evaluate the integral

Evaluate the indefinite integral `int x(8x+7)^8 dx`

`int (x^3 + 2x^2 + 3x - 2)/(x^2 + 2x + 2)^2 dx` Evaluate the integral

`int x/sqrt(x^2 + x + 1) dx` Evaluate the integral