What is the integral of (x^4+8)/(x^3) ?

The integral of (x^4+8)/(x^3) is (x^2)/2-4/(x^2)+C

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integral of (x^4+8)/(x^3)

\int \frac{x ^{4} + 8}{x ^{3}} d x

\frac{x ^{2}}{2} - \frac{4}{x ^{2}} + C

Solution steps

\int \frac{x ^{4} + 8}{x ^{3}} d x

Expand

\frac{x ^{4} + 8}{x ^{3}} : x + \frac{8}{x ^{3}}

= \int x + \frac{8}{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 \frac{8}{x ^{3}} d x

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

\int \frac{8}{x ^{3}} d x = - \frac{4}{x ^{2}}

= \frac{x ^{2}}{2} - \frac{4}{x ^{2}}

Add a constant to the solution

= \frac{x ^{2}}{2} - \frac{4}{x ^{2}} + C

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What is the integral of (x^4+8)/(x^3) ?
The integral of (x^4+8)/(x^3) is (x^2)/2-4/(x^2)+C