What is the integral of 4-6x-40x^3 ?

The integral of 4-6x-40x^3 is 4x-3x^2-10x^4+C

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integral of 4-6x-40x^3

\int 4 - 6 x - 40 x^{3} d x

4 x - 3 x^{2} - 10 x^{4} + C

Solution steps

\int 4 - 6 x - 40 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 4 d x - \int 6 x d x - \int 40 x^{3} d x

\int 4 d x = 4 x

\int 6 x d x = 3 x^{2}

\int 40 x^{3} d x = 10 x^{4}

= 4 x - 3 x^{2} - 10 x^{4}

Add a constant to the solution

= 4 x - 3 x^{2} - 10 x^{4} + C

\int_{0}^{2} 1 + e^{- 2 x} d x

n = 1 \sum \infty \frac{3}{( - 2 ) ^{n}}

\int_{0}^{\infty} \frac{1}{5 + x} d x

f (x) = x^{2}

5 arcsin (\frac{x - 1}{x + 1})

What is the integral of 4-6x-40x^3 ?
The integral of 4-6x-40x^3 is 4x-3x^2-10x^4+C