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

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

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

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

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

Solution steps

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

Expand

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

= \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{d}{d x} (\frac{3}{x} + 5)

\int (\frac{1}{x ln ( x )}) d x

\frac{\partial}{\partial x} (x y^{2} - 5 y + 6)

f (x) = 6 x sin (x), at x = \frac{π}{2}

\frac{d}{d x} (c e^{4 x})

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