What is the integral of (x^4+2x^3-3)/(x^3) ?

The integral of (x^4+2x^3-3)/(x^3) is (x^2)/2+2x+3/(2x^2)+C

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integral of (x^4+2x^3-3)/(x^3)

\int \frac{x ^{4} + 2 x ^{3} - 3}{x ^{3}} d x

\frac{x ^{2}}{2} + 2 x + \frac{3}{2 x ^{2}} + C

Solution steps

\int \frac{x ^{4} + 2 x ^{3} - 3}{x ^{3}} d x

Expand

\frac{x ^{4} + 2 x ^{3} - 3}{x ^{3}} : x + 2 - \frac{3}{x ^{3}}

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

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

\int 2 d x = 2 x

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

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

Simplify

= \frac{x ^{2}}{2} + 2 x + \frac{3}{2 x ^{2}}

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + \frac{3}{2 x ^{2}} + C

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

5 x y (\frac{d y}{d x}) = 1

\frac{d}{d x} (3 - 7 x - x^{2})

\int \frac{x ^{3}}{x ^{4} + 4} d x

\frac{s}{s ^{2} + s - 2}

What is the integral of (x^4+2x^3-3)/(x^3) ?
The integral of (x^4+2x^3-3)/(x^3) is (x^2)/2+2x+3/(2x^2)+C