1) First step is to factorize the denominator of your fraction completely:
x^2 + 4x - 5 = (x + 5)(x - 1)x2+4x−5=(x+5)(x−1)
In case you don't know how to do this:
- solve the quadratic equation x^2 + 4x - 5 = 0x2+4x−5=0
- find the solutions x = -5x=−5 and x = 1x=1
- your factorization is (x - "solution"_1)(x - "solution"_2)(x−solution1)(x−solution2)
2) Now that you have the factorization, you need to decompose your fraction into partial fractions.
This means that you are searching for AA and BB so that
(x+4)/((x+5)(x-1)) = A/(x+5) + B/(x-1)x+4(x+5)(x−1)=Ax+5+Bx−1
... multiply both sides of the equation with (x+5)(x-1)(x+5)(x−1)...
<=> x + 4 = A(x-1) + B(x+5)⇔x+4=A(x−1)+B(x+5)
<=> x + 4 = A * x - A + B * x + 5B⇔x+4=A⋅x−A+B⋅x+5B
<=> color(blue)(x) + color(red)(4) = color(blue)(A * x) color(white)(x) color(red)(- A) + color(blue)(B * x) + color(red)(5B)⇔x+4=A⋅xx−A+B⋅x+5B
To solve this equation, you need to compare the color(blue) xx terms and the color(red)"constant"constant terms:
{ (1 = A + B), (4 = - A + 5B) :}
The solution of this linear equation system is B = 5/6, A = 1/6.
This means that you have succeded in the partial fraction decomposition:
(x+4)/((x+5)(x-1)) = 1/6 * 1/(x+5) + 5/6 * 1/(x-1)
3) The only thing left to do is solving the integral. :-)
int (x+4)/(x^2 + 4x - 5) "d"x = int (x+4)/ ((x+5)(x-1)) "d"x
color(white)(xxxxxxxxxxx) = int 1/6 * 1/(x+5) + 5/6 * 1/(x-1) "d"x
color(white)(xxxxxxxxxxx) = 1/6 * int 1/(x+5) "d"x + 5/6 * int 1/(x-1) "d"x
color(white)(xxxxxxxxxxx) = 1/6 * ln |x+5| + 5/6 * ln | x-1 |
