What is the integral of (x^4-4x^{-3}-5) ?

The integral of (x^4-4x^{-3}-5) is (x^5)/5+2/(x^2)-5x+C

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integral of (x^4-4x^{-3}-5)

\int (x^{4} - 4 x^{- 3} - 5) d x

\frac{x ^{5}}{5} + \frac{2}{x ^{2}} - 5 x + C

Solution steps

\int x^{4} - 4 x^{- 3} - 5 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^{4} d x - \int 4 x^{- 3} d x - \int 5 d x

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int 5 d x = 5 x

= \frac{x ^{5}}{5} - (- \frac{2}{x ^{2}}) - 5 x

Simplify

= \frac{x ^{5}}{5} + \frac{2}{x ^{2}} - 5 x

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{2}{x ^{2}} - 5 x + C

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

\int \frac{1}{x ( 5 - x )} d x

\int \frac{ln ( x + 9 )}{x + 9} d x

\int_{- 1}^{9} (9 - x^{2}) - (- 8 x) d x

y^{^{'}} = - 3 y^{2}

What is the integral of (x^4-4x^{-3}-5) ?
The integral of (x^4-4x^{-3}-5) is (x^5)/5+2/(x^2)-5x+C