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
Explanation:
First we should get in the numerator a polynomial of a grade inferior than the denominator's
By long division:
_____
So the expression becomes
Let's deal with the last part
Trying
By long division:
___
So we can write the last integrand as
For
-45/26=A/-2+C/13
-29/18=-A+(B+C)/18
27/34=A+(3B+C)/34
Or
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
Let's deal with the last part
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
Butsin y=(x+2)/3cosy
Andsin^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