What is the integral of (3/(x^2)-1) ?

The integral of (3/(x^2)-1) is -3/x-x+C

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integral of (3/(x^2)-1)

\int (\frac{3}{x ^{2}} - 1) d x

- \frac{3}{x} - x + C

Solution steps

\int \frac{3}{x ^{2}} - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{3}{x ^{2}} d x - \int 1 d x

\int \frac{3}{x ^{2}} d x = - \frac{3}{x}

\int 1 d x = x

= - \frac{3}{x} - x

Add a constant to the solution

= - \frac{3}{x} - x + C

\frac{d y}{d x} = 2 - y

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 2

\int \frac{1}{9 x ^{2} + 5} d x

144 y^{^{'''}} + 120 y^{^{''}} + 25 y^{^{'}} = 0

\int (4 x + 7) e^{x} d x