Question

How do you evaluate `\int \frac { ( x - 2) } { x ( x ^ { 2} - 4x + 5) ^ { 2} } d x`?

Answer 1

`int \ (x-2)/(x(x^2-4x+5)^2) \ dx = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c`

Explanation:

Let us denote the required integral by `I`

`I = int \ (x-2)/(x(x^2-4x+5)^2) \ dx`

Partial Fraction Decomposition

We can decompose the integrand using partial fractions, the partial fraction decomposition will be of the form:

`(x-2)/(x(x^2-4x+5)^2) = A/x + (Bx+C)/(x^2-4x+5) + (Dx+E)/(x^2-4x+5)^2`
`" " = {A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x}/{x(x^2-4x+5)^2}`

Leading to:

`x-2 = A(x^2-4x+5)^2 + (Bx+C)x(x^2-4x+5) + (Dx+E)x`

`:. x-2 = A(x^4-8x^3+26x^2-40x+25) + (Bx+C)(x^3-4x^2+5x) + (Dx+E)x`

We can use various methods to find our unknown constants:

`"Put " x=0 => -2=A*5^2`
`:. A = -2/25`

Compare coefficients:

`"Coeff"(x^4) => 0 = A +B`
`:. B = 2/25`

`"Coeff"(x^3) => 0 = -8A+C-4B`
`:. 0=16/25+C-8/25 => C=-8/25`

`"Coeff"(x^2) => 0 = 26A+5B-4C+D`
`:. 0=-52/25+10/25+32/25+D => D=10/25=2/5`

`"Coeff"(x^1) => 1 = -40A+5C+E`
`:. 1=80/25-40/25+E => E=-3/5`

Hence we can the integral as:

`I = int \ (-2/25)/x + (2/25x-8/25)/(x^2-4x+5) + (2/5x-3/5)/(x^2-4x+5)^2 \ dx`

`\ \ = -2/35 \ int \ 1/x dx + 2/25 int (x-4)/(x^2-4x+5) \ dx+ 1/5 int (2x-3)/(x^2-4x+5)^2 \ dx`

`\ \ = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3`

Where:

`I_1 = int \ 1/x dx + 2/25 \ dx`
`I_2 = int (x-4)/(x^2-4x+5) \ dx`
`I_3 = int (2x-3)/(x^2-4x+5)^2 \ dx`

Now let us take each of these three separate integrals in turn:

Integral 1: `I_1`

The first integral, `I_1`, we can just evaluate directly:

`I_1 = int \ 1/x \ dx`
`\ \ \ = ln|x|`

Integral 2: `I_2`

The second integral, `I_2`, we manipulate by completing the square:

`I_2 = int \ (x-4)/(x^2-4x+5) \ dx`
`\ \ \ = int \ (x-4)/((x-2)^2-2^2+5) \ dx`
`\ \ \ = int \ (x-4)/((x-2)^2+1) \ dx`

Let `u=x-2 => (du)/dx = 1`, and `x-4=u-2`, then substituting gives

`I_2 = int \ (u-2)/(u^2+1) \ du`
`\ \ \ = int \ u/(u^2+1) - 2/(u^2+1)\ du`
`\ \ \ = 1/2int \ (2u)/(u^2+1) \ du - 2 int \ 1/(u^2+1)\ du`
`\ \ \ = 1/2ln|u^2+1| - 2 arctanu`

Restoring the substitution we get:

`I_2 = 1/2ln|x^2-4x+5| - 2 arctan(x-2)`

Integral 3: `I_3`

The third integral, `I_3` we continue as earlier by completing the square to get:

`I_3 = int \ (2x-3)/(x^2-4x+5)^2 \ dx`
`\ \ \ = int (2x-3)/((x-2)^2+1)^2 \ dx`

Let `u=x-2 => (du)/dx = 1`, and `2x-3=2u+1`, then substituting gives

`I_3 = int (2u+1)/(u^2+1)^2 \ du`
`\ \ \ = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du`

Consider the integral:

`int (2u)/(u^2+1)^2 \ du`

Let `w=u^2+1 => (dw)/(du)=2u`, so we can substitute to get:

`int (2u)/(u^2+1)^2 \ du = int \ 1/w^2 \ dw`
`" " = -1/w`

Restoring the substitutions we get:

`int (2u)/(u^2+1)^2 \ du = -1/(u^2+1)`
`" " = -1/((x-2)^2+1)`
`" " = -1/(x^2-4x+5)`

And now we consider:

`int 1/(u^2+1)^2 \ du`

Let `u=tan theta => (du)/(d theta) = sec^2 theta`, then substituting gives

`int 1/(u^2+1)^2 \ du = int \ 1/(tan^2theta+1)^2 \ sec^2 theta \ d theta`
`" " = int \ 1/(sec^2theta)^2 \ sec^2 theta \ d theta`
`" " = int \ 1/(sec^2theta) \ d theta`
`" " = int \ cos^2theta \ d theta`
`" " = int \ 1/2(1+cos2theta) \ d theta`
`" " = 1/2 \ int \ 1+cos2theta \ d theta`
`" " = 1/2 (theta + 1/2 sin2theta)`
`" " = 1/2 theta + 1/2 sintheta costheta`

Now, `u=tan theta => u=sintheta/costheta`

`:. sinthetacostheta = cos^2theta * u`
`" " =1/sec^2theta * u`
`" " =1/(1+tan^2theta) * u`
`" " =u/(u^2+1)`

Restoring the substitutions we get:

`int 1/(u^2+1)^2 \ du = 1/2 arctanu+ 1/2 u/(u^2+1)`
`" " = 1/2 arctan(x-2) + 1/2 (x-2)/((x-2)^2+1)`
`" " = 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)`

Hence, we can write the third integral as:

`I_3 = int (2u)/(u^2+1)^2 \ du+ int 1/(u^2+1)^2 \ du`
`\ \ \ = -1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)`

Combine Results

Now we return back to our original partial fraction result from earlier, and replace the three integrals with the above results:

`I = -2/25 \ I_1 + 2/25 \ I_2 + 1/5 \ I_3`

`\ \ = -2/25 {ln|x|} + 2/25 {1/2ln|x^2-4x+5| - 2 arctan(x-2)} + 1/5 {-1/(x^2-4x+5)+ 1/2 arctan(x-2) + 1/2 (x-2)/(x^2-4x+5)} + c`

`\ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 4/25 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 arctan(x-2) + 1/10 (x-2)/(x^2-4x+5) + c`

`\ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) -1/5 1/(x^2-4x+5)+ 1/10 (x-2)/(x^2-4x+5) + c`

`\ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-2-2)/(10(x^2-4x+5)) + c`

`\ \ = -2/25 ln|x| + 1/25 ln|x^2-4x+5| - 3/50 arctan(x-2) + (x-4)/(10(x^2-4x+5)) + c`