Solve the differential equation

f''(x)=2/x^2, f'(1)=4, f(1)=3

The first one is the 2nd derivative.

So far I got the anti-derivative -2\x + c

and then f'(1)= -2/(1) + 4 =2 so, -2\x+2 dx

but now I'm lost

f"(x) = 2x⁻²

Integrate to get
f'(x) = -2/x + A
f'(1) = -2 + A = 4 >>> A = 6
f'(x) = -2/x + 6

Integrate again to get
f(x) = -2ln(x) + 6x + B
f(1) = 6 + B = 3 >>> B = -3

So our final answer is
f(x) = 6x - 2ln(x) - 3

f '(x) = -2/x + C

4 = C - 2 --> C = 6

f '(x) = -2/x + 6

f(x) = -2*ln|x| + 6x + D

3 = D + 6 --> D = -3

f(x) = -2*ln(x) + 6x - 3