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

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

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

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

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

Solution steps

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

\int 4 d x = 4 x

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

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

= 4 x - \frac{x ^{4}}{4} - \frac{5 x ^{6}}{6}

Add a constant to the solution

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

x \to 0 + lim (tan^{x} (6 x))

x \to 0 lim (e^{1})

\int x^{3} x^{2} + 44 d x

x \to 0 lim (\frac{( cos ( a x ) - cos ( b x ) )}{x ^{2}})

\frac{d}{d x} (\frac{2 x ^{2}}{cos ( x )})

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