What is the integral of (x^5+4x^4-3x^{-2}+x-1) ?

The integral of (x^5+4x^4-3x^{-2}+x-1) is (x^6)/6+(4x^5)/5+3/x+(x^2)/2-x+C

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integral of (x^5+4x^4-3x^{-2}+x-1)

\int (x^{5} + 4 x^{4} - 3 x^{- 2} + x - 1) d x

\frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} + \frac{3}{x} + \frac{x ^{2}}{2} - x + C

Solution steps

\int x^{5} + 4 x^{4} - 3 x^{- 2} + x - 1 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^{5} d x + \int 4 x^{4} d x - \int 3 x^{- 2} d x + \int x d x - \int 1 d x

\int x^{5} d x = \frac{x ^{6}}{6}

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

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

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

\int 1 d x = x

= \frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} - (- \frac{3}{x}) + \frac{x ^{2}}{2} - x

Simplify

= \frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} + \frac{3}{x} + \frac{x ^{2}}{2} - x

Add a constant to the solution

= \frac{x ^{6}}{6} + \frac{4 x ^{5}}{5} + \frac{3}{x} + \frac{x ^{2}}{2} - x + C

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What is the integral of (x^5+4x^4-3x^{-2}+x-1) ?
The integral of (x^5+4x^4-3x^{-2}+x-1) is (x^6)/6+(4x^5)/5+3/x+(x^2)/2-x+C