What is the integral of (x^{1.8}+9x^{3.5}) ?

The integral of (x^{1.8}+9x^{3.5}) is 0.35714…x^{2.8}+2x^{4.5}+C

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integral of (x^{1.8}+9x^{3.5})

\int (x^{1.8} + 9 x^{3.5}) d x

0.35714 \dots x^{2.8} + 2 x^{4.5} + C

Solution steps

\int x^{1.8} + 9 x^{3.5} 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 x^{1.8} d x + \int 9 x^{3.5} d x

\int x^{1.8} d x = 0.35714 \dots x^{2.8}

\int 9 x^{3.5} d x = 2 x^{4.5}

= 0.35714 \dots x^{2.8} + 2 x^{4.5}

Add a constant to the solution

= 0.35714 \dots x^{2.8} + 2 x^{4.5} + C

(4 y + x^{2} y) d y - (2 x + x y^{2}) d x = 0

y^{^{''}} - y^{^{'}} - 2 y = 8 e^{- t}

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

4 x^{- 1}

\frac{\partial}{\partial y} (x^{2} - 4 y^{2})

What is the integral of (x^{1.8}+9x^{3.5}) ?
The integral of (x^{1.8}+9x^{3.5}) is 0.35714…x^{2.8}+2x^{4.5}+C