What is the integral of (x^{1.1}+9x^{3.5}) ?

The integral of (x^{1.1}+9x^{3.5}) is 0.47619…x^{2.1}+2x^{4.5}+C

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integral of (x^{1.1}+9x^{3.5})

\int (x^{1.1} + 9 x^{3.5}) d x

0.47619 \dots x^{2.1} + 2 x^{4.5} + C

Solution steps

\int x^{1.1} + 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^{1.1} d x + \int 9 x^{3.5} d x

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

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

= 0.47619 \dots x^{2.1} + 2 x^{4.5}

Add a constant to the solution

= 0.47619 \dots x^{2.1} + 2 x^{4.5} + C

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What is the integral of (x^{1.1}+9x^{3.5}) ?
The integral of (x^{1.1}+9x^{3.5}) is 0.47619…x^{2.1}+2x^{4.5}+C