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

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

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integral of x+1/(x^2)

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

\frac{x ^{2}}{2} - \frac{1}{x} + C

Solution steps

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

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

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

= \frac{x ^{2}}{2} - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{1}{x} + C

\frac{\partial}{\partial x} (x y - 1)

\frac{\partial}{\partial y} ((x^{2} - 1) (y + 2))

f (x) = \frac{3}{x ^{2}}, at x = 1

\int_{0}^{\frac{π}{4}} 4 sin^{2} (x) d x

h \to 0 lim ((1 + 5 h)^{\frac{1}{h}})