What is the integral of (x^{1.1}+7x^{2.5}) ?

The integral of (x^{1.1}+7x^{2.5}) is 0.47619…x^{2.1}+2x^{3.5}+C

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integral of (x^{1.1}+7x^{2.5})

\int (x^{1.1} + 7 x^{2.5}) d x

0.47619 \dots x^{2.1} + 2 x^{3.5} + C

Solution steps

\int x^{1.1} + 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^{1.1} d x + \int 7 x^{2.5} d x

\int x^{1.1} d x = 0.47619 \dots x^{2.1}

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

= 0.47619 \dots x^{2.1} + 2 x^{3.5}

Add a constant to the solution

= 0.47619 \dots x^{2.1} + 2 x^{3.5} + C

x \to 2 lim (3 x^{3} - 1)

\frac{\partial}{\partial t} (tan (t))

(2 x) y^{^{'}} = x + 3 y

s + 2

y^{^{''}} - 6 y^{^{'}} + 9 y = 30 t^{2} e^{3 t}

What is the integral of (x^{1.1}+7x^{2.5}) ?
The integral of (x^{1.1}+7x^{2.5}) is 0.47619…x^{2.1}+2x^{3.5}+C