What is the integral of (x^{1.4}+7x^{2.5}) ?

The integral of (x^{1.4}+7x^{2.5}) is 0.41666…x^{2.4}+2x^{3.5}+C

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integral of (x^{1.4}+7x^{2.5})

\int (x^{1.4} + 7 x^{2.5}) d x

0.41666 \dots x^{2.4} + 2 x^{3.5} + C

Solution steps

\int x^{1.4} + 7 x^{2.5} 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^{1.4} d x + \int 7 x^{2.5} d x

\int x^{1.4} d x = 0.41666 \dots x^{2.4}

\int 7 x^{2.5} d x = 2 x^{3.5}

= 0.41666 \dots x^{2.4} + 2 x^{3.5}

Add a constant to the solution

= 0.41666 \dots x^{2.4} + 2 x^{3.5} + C

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

f (x) = \frac{x}{x ^{2} + 4}

t^{3} y^{^{'}} + 4 t^{2} y = e^{- t}, y (- 1) = 0

\int - 3 sin (x) d x

x (x^{2} + 1)^{10}

What is the integral of (x^{1.4}+7x^{2.5}) ?
The integral of (x^{1.4}+7x^{2.5}) is 0.41666…x^{2.4}+2x^{3.5}+C