Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 7

int 5/(x^2+3x-4)dx
To solve using partial fraction, the denominator of the integrand should be factored.
5/(x^2+3x-4)=5/((x+4)(x-1))
Then, express it as sum of fractions.
5/((x+4)(x-1))=A/(x+4)+B/(x-1)
To determine the values of A and B, multiply both sides by the LCD of the fractions present.
(x+4)(x-1)*5/((x+4)(x-1))=(A/(x+4)+B/(x-1))*(x+4)(x-1)
5=A(x-1)+B(x+4)
Then, assign values to x in which either x+4 or x-1 will become zero.
So plug-in x=-4 to get the value of A.
5=A(-4-1)+B(-4+4)
5=A(-5)+B(0)
5=-5A
-1=A
Also, plug-in x=1
5=A(1-1)+B(1+4)
5=A(0)+B(5)
5=5B
1=B
So the partial fraction decomposition of the integrand is
int 5/(x^2+3x-4)dx
= int 5/((x+4)(x-1))dx
= int (-1/(x+4)+1/(x-1))dx
Then, express it as two integrals.
= int -1/(x+4)dx + int 1/(x-1)dx
= - int 1/(x+4)+int 1/(x-1)dx
To take the integral, apply the formula int 1/u du = ln|u| + C .
= -ln|x+4| + ln|x-1| + C

Therefore, int 5/(x^2+3x-4)dx= -ln|x+4| + ln|x-1| + C .