How do you integrate #int (x+4)/(x^2+4x-5) dx# using partial fractions?

1) First step is to factorize the denominator of your fraction completely:

#x^2 + 4x - 5 = (x + 5)(x - 1)#

In case you don't know how to do this:
- solve the quadratic equation #x^2 + 4x - 5 = 0#
- find the solutions #x = -5# and #x = 1#
- your factorization is #(x - "solution"_1)(x - "solution"_2)#

2) Now that you have the factorization, you need to decompose your fraction into partial fractions.

This means that you are searching for #A# and #B# so that

#(x+4)/((x+5)(x-1)) = A/(x+5) + B/(x-1)#

... multiply both sides of the equation with #(x+5)(x-1)#...

#<=> x + 4 = A(x-1) + B(x+5)#
#<=> 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)#

To solve this equation, you need to compare the #color(blue) x# terms and the #color(red)"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 |#