calculus problems?

Question

Find f.
f ''(x) = x^-2
x > 0
f(1) = 0
f(3) = 0

A particle is moving with the given data. Find the position of the particle.
a(t) = t - 4
s(0) = 1
v(0) = 5

Find f.
f ''(x) = 24x^2 + 2x + 2
f(1) = 2
f '(1) = -2

Answer 1

1)

int(x^-2)dx = x^-1 + c
-int(x^-1+c)dx =-ln(x) + cx + d

-ln(1)+c+d=-ln(3)+3c+d
c=ln(3)/2
d=-c=-ln(3)/2

Answer: f(x) = -ln(x) + (ln(3)/2)(x-1)

I don't get the second one

3)
int(24x^2+2x+2)dx = 8x^3+x^2+2x+c
int(8x^3+x^2+2x+c)dx = 2x^4+(x^3)/3+x^2+cx+d

8+1+2+c=2
c=-9

2+1/3+1-9+d=-2
d= 11/3

Answer: 2x^4+(x^3)/3+x^2-2x+11/3