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)

integral

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

\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})

Simplify

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

Add a constant to the solution

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

\frac{z}{( 1 - z ^{2} ) ^{\frac{1}{2}}}

3 x \cdot e^{1.5 x^{2}}

e^{x} \cdot cos^{2} (e^{x})

\frac{e ^{- 1}}{x ^{3}}

\frac{4}{x ^{3}}

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