How do you evaluate the integral of int ( (2x^4 + 5x^3 + 17x^2 - 42x + 19) / (x^3 + 2x^2 + 5x - 26) ) dx?

1 Answer
Mar 8, 2016

=x^2+x+23/25ln| x-2|-102/25ln|3/sqrt(x^2+4x+13)|-41/5tan^(-1)((x+2)/3)+const.

Explanation:

First we should get in the numerator a polynomial of a grade inferior than the denominator's

By long division:

" "2x^4+5x^3+0x^2+2x-1" "|" "x^3+2x^2+5x-26
-2x^4-4x^3-10x^2+26x" "|____
_____" "2x+1
" "x^3+7x^2-16x+19
" "-x^3-2x^2-5x+26
" " _______
" "5x^2-21x+45

So the expression becomes
=int(2x+1)dx+int (5x^2-21x+45)/(x^3+2x^2+5x-26)dx

Let's deal with the last part
Trying +-1, +-2 and +-3 we discover that x=2 is a root of the polynomial of the denominator
By long division:

" "x^3+2x^2+5x-26" "|" "x-2
-x^3+2x^2" "|____
___" "x^2+4x+13
" "4x^2+5x
" "-4x^2+8x
" "
___
" "13x-26
" "-13x+26
" "
______
" "0

So we can write the last integrand as
(5x^2-21x+45)/((x-2)(x^2+4x+13))=A/(x-2)+(Bx+C)/(x^2+4x+13)

For x=0, 1 and 3 we get

-45/26=A/-2+C/13
-29/18=-A+(B+C)/18
27/34=A+(3B+C)/34

Or
[[-1/2,0,1/13],[-1,1/18,1/18],[1, 3/34,1/34]][[A],[B],[C]]=[[-45/26],[-29/18],[27/34]]
Solving this system of variables, we get

A=.92=23/25
B=4.08=102/25
C=-16.52=-413/25

Then the original expression becomes
=x^2+x+23/25int(x-2)dx+1/25int(102x-413)/(x^2+4x+13)dx

Let's deal with the last part
int(102x-413)/(x^2+4x+13)dx=

Since (x+2)^2=x^2+4x+4
(x+2)=3tany
dx=3sec^2y*dy
How many units of (x+2) are there in the numerator?
-> (102x-413)/(x+2)=102-615/(x+2)
So the last partial expression becomes
=102int (3tany.3cancel(sec^2 y))/(9cancel(sec^2y))dy-615int (3cancel(sec^2 y))/(9cancel(sec^y))dy
=-102ln|cosy|-205y
But sin y=(x+2)/3cosy
And sin^2y+cos^2y=1 => ((x^2+4x+4)/9+1)cos^2 y=1 => cosy=3/sqrt(x^2+4x+13)
So the partial expression becomes
=-102ln|3/sqrt(x^2+4x+13)|-205tan^(-1)((x+2)/3)

Finally, back to the main expression, we get
=x^2+x+23/25ln| x-2|+1/25(-102ln|3/sqrt(x^2+4x+13)|-205tan^(-1)((x+2)/3))
=x^2+x+23/25ln| x-2|-102/25ln|3/sqrt(x^2+4x+13)|-41/5tan^(-1)((x+2)/3)+const.