calculus
posted by travis on .
f''(x) = (3x^3  5x^2 + 1) / x^2; and f(1) = 0, f(1) = 1.
So i turn f''(x) into (3x^3  5x^2 + 1)(x^2) and multiply to get f''(x) = 3x  5 + x^2
So f'(x) = (3/2)x^2  5x  (1/x) + c
And f(x) = (1/2)x^3  (5/2)x^2  ln(x) + cx + d
Not really sure what to do from here, and when plugging in 1 to ln(x) I encounter a problem. Thanks for any help

The math looks odd, but the answer is correct as far as you go.
Now go back to your boundary conditions, f(1) and f(1)
I think you better change ln(x) to lnx while you're at it, or ln(1) is not defined.
0 = 1/2  5/2  0 + c + d
1 = (1/2)  5/2  0  c + d
c = 1, d=3, so
f(x) = 1/2 x^3  5/2 x^2  x + 3  lnx