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

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

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integral of (4x^4-3x^{-2}+5x^{-5}+2)

\int (4 x^{4} - 3 x^{- 2} + 5 x^{- 5} + 2) d x

\frac{4 x ^{5}}{5} + \frac{3}{x} - \frac{5}{4 x ^{4}} + 2 x + C

Solution steps

\int 4 x^{4} - 3 x^{- 2} + 5 x^{- 5} + 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 4 x^{4} d x - \int 3 x^{- 2} d x + \int 5 x^{- 5} d x + \int 2 d x

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

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

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

\int 2 d x = 2 x

= \frac{4 x ^{5}}{5} - (- \frac{3}{x}) - \frac{5}{4 x ^{4}} + 2 x

Simplify

= \frac{4 x ^{5}}{5} + \frac{3}{x} - \frac{5}{4 x ^{4}} + 2 x

Add a constant to the solution

= \frac{4 x ^{5}}{5} + \frac{3}{x} - \frac{5}{4 x ^{4}} + 2 x + C

\int x^{2} (5 + x)^{2} d x

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

\int (\frac{x ^{3}}{x ^{4} + 9}) d x

\int [x (x^{2} - 16)^{2} - \frac{1}{2 x + 5}] d x

\int \frac{3 x + 5}{x ^{2} + 4} d x

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