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

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

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

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

\frac{1}{3} (\frac{x ^{2}}{2} - \frac{1}{x}) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

\frac{x ^{3} + 1}{x ^{2}} : x + \frac{1}{x ^{2}}

= \frac{1}{3} \cdot \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

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

Add a constant to the solution

= \frac{1}{3} (\frac{x ^{2}}{2} - \frac{1}{x}) + C

\frac{\partial}{\partial x} (8 x cos (5 x y))

x \to 0 + lim (\frac{0}{x})

\frac{d y}{d x} = \frac{1}{y + x}

f (x) = x^{2} + 7, at x = 3

\frac{\partial}{\partial y} (- 3 x e^{- x y})

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