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

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

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

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

\frac{4 x ^{5}}{5} + x^{3} + x^{2} + x - \frac{1}{4 x ^{4}} + C

Solution steps

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

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

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

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

\int 1 d x = x

\int x^{- 5} d x = - \frac{1}{4 x ^{4}}

= \frac{4 x ^{5}}{5} + x^{3} + x^{2} + x - \frac{1}{4 x ^{4}}

Add a constant to the solution

= \frac{4 x ^{5}}{5} + x^{3} + x^{2} + x - \frac{1}{4 x ^{4}} + C

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