f'(x) = (5x-3)/(x^2+4)

So f(x) is the Int (5x-3)/(x^2+4)dx +C

f(x) = Int(5x-3)/(x^2+4) dx= Int {5/2 (2x) dx-3Int{dx/x^2+4 }'...(1)

We do the two integrations on the right separately :

Int (5/2) 2xdx/(x^2+4) = In (5/2) dt/t , whre t = 1/(x^2+4).

= (5/2)logt = (5/2)ln (x^2+4).

Int(5/2)(2x)dx/(x^2+1) =(5/2) ln(x^2+4)

-3Int {dx/(x^2+4) = -3 (1/2) arc tan (x/2).

Therefore , from (1) we get:

f(x) = (5/2)ln(x^2+4) -(3/a) arctan (x/2) +C.

To determine the function f(x), we'll have to integrate (5x-3)/(x^2+4)

Int (5x-3)dx/(x^2+4) = f(x) + C

We'll use the additive property of the integral:

Int (5x-3)dx/(x^2+4) = Int 5xdx/(x^2+4) - Int 3dx/(x^2+4)

We'll note Int 5xdx/(x^2+4) = 5 I1

Int 3dx/(x^2+4) = 3I2

We'll calculate I1 using substitution technique.

We'll note x^2+4 = t

We'll differentiate both sides:

2xdx = dt

We'll substitute in the original integral and we'll get:

I1 = Int dt/t

I1 = ln |t| + C, but t = x^2+4 > 0

I1 = ln (x^2+4) + C

5I1 = 5ln (x^2+4) + C

**5I1 = ln (x^2+4)^5 + C**

We'll calculate I2:

I2 = Int dx/(x^2+4)

I2 = [arctan (x/2)]/2 + C

**3I2 = 3[arctan (x/2)]/2 + C**

**Int (5x-3)dx/(x^2+4) = ln (x^2+4)^5 + 3[arctan (x/2)]/2 + C**