`int 5/(x^2+3x-4) dx` Use partial fractions to find the indefinite integral

`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)`

...

`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` .