Integral of (x+2)/(x^2+3x-4)

I need a step by step answer please, this is an example I'm working on to study for a midterm.

Note that the denominator factors to (x + 4)(x - 1), so, by Partial Fractions, we can write:
(x + 2)/(x^2 + 3x - 4) = A/(x + 4) + B/(x - 1), for some A and B.

Multiplying both sides by x^2 + 3x - 4 gives:
x + 2 = A(x - 1) + B(x + 4).

(i) Letting x = -4 ==> -5A = -2 ==> A = 2/5
(ii) Letting x = 1 ==> 5B = 3 ==> B = 3/5.

Thus:
(x + 2)/(x^2 + 3x - 4) = (2/5)/(x + 4) + (3/5)/(x - 1).

Integrating term-by-term now yields:
∫ (x + 2)/(x^2 + 3x - 4) dx
= ∫ [(2/5)/(x + 4) + (3/5)/(x - 1)] dx
= (2/5)ln|x + 4| + (3/5)ln|x - 1| + C.

I hope this helps!

We can break this down into partial fractions:

(x + 2) / (x^2 + 3x - 4) =>
(x + 2) / ((x + 4) * (x - 1)) =>
A / (x + 4) + B / (x - 1)

A * (x - 1) + B * (x + 4) = (x + 2)
Ax - A + Bx + 4B = x + 2

Ax + Bx = x
-A + 4B = 2

A + B = 1
-A + 4B = 2

A + B - A + 4B = 1 + 2
5B = 3
B = (3/5)

A + B = 1
A + (3/5) = 1
A = 2/5

(2/5) / (x + 4) + (3/5) / (x - 1)

Now integrate:

(2/5) * ln(x + 4) + (3/5) * ln(x - 1) + C
(1/5) * (2 * ln(x + 4) + 3 * ln(x - 1)) + C
(1/5) * (ln((x + 4)^2) + ln((x - 1)^3)) + C
(1/5) * ln( (x + 4)^2 * (x - 1)^3 ) + C

∫(x + 2)/(x² + 3x - 4)

∫(x + 2)/(x + 4)(x - 1) dx

∫(x - 1 + 3)/(x + 4)(x - 1) dx

∫dx/(x + 4) + 3/5∫[(x + 4) - (x - 1)]/(x + 4)(x - 1) dx

ln|x + 4| + 3/5*ln|(x - 1)/(x + 4)| + C