What is the solution for f^{''}(x)=(x^2-2x+1)/(x^2) ?

The solution for f^{''}(x)=(x^2-2x+1)/(x^2) is f(x)=(x^2)/2-2(xln(x)-x)-ln(x)+c_{1}x+c_{2}

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f^{''}(x)=(x^2-2x+1)/(x^2)

f^{^{''}} (x) = \frac{x ^{2} - 2 x + 1}{x ^{2}}

f (x) = \frac{x ^{2}}{2} - 2 (x ln (x) - x) - ln (x) + c_{1} x + c_{2}

Solution steps

f^{''} (x) = \frac{x ^{2} - 2 x + 1}{x ^{2}}

Solve

f^{''} (x) = \frac{x ^{2} - 2 x + 1}{x ^{2}} : f^{'} (x) = x - 2 ln (x) - \frac{1}{x} + c_{1}

Solve

f^{'} (x) = x - 2 ln (x) - \frac{1}{x} + c_{1} : f (x) = \frac{x ^{2}}{2} - 2 (x ln (x) - x) - ln (x) + c_{1} x + c_{2}

f (x) = \frac{x ^{2}}{2} - 2 (x ln (x) - x) - ln (x) + c_{1} x + c_{2}

\frac{d}{d x} (x - x^{3} sin (x))

\frac{s}{( s ^{2} + 1 ) ( s ^{2} + 2 s + 2 )}

\frac{d y}{d x} = y (x y^{6} + 3)

16 x, x^{3}, [- 4, 4]

\frac{d y}{d x} = (x + y + 6)^{2}

What is the solution for f^{''}(x)=(x^2-2x+1)/(x^2) ?
The solution for f^{''}(x)=(x^2-2x+1)/(x^2) is f(x)=(x^2)/2-2(xln(x)-x)-ln(x)+c_{1}x+c_{2}