What is the integral of x\sqrt[9]{1+x^2} ?

The integral of x\sqrt[9]{1+x^2} is 9/20 (1+x^2)^{10/9}+C

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integral of x\sqrt[9]{1+x^2}

\frac{9}{20} (1 + x^{2})^{\frac{10}{9}} + C

Solution steps

Apply u-substitution

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

= \frac{1}{2} \cdot \frac{9}{10} u^{\frac{10}{9}}

Substitute back

u = 1 + x^{2}

= \frac{1}{2} \cdot \frac{9}{10} (1 + x^{2})^{\frac{10}{9}}

Simplify

\frac{1}{2} \cdot \frac{9}{10} (1 + x^{2})^{\frac{10}{9}} : \frac{9}{20} (1 + x^{2})^{\frac{10}{9}}

= \frac{9}{20} (1 + x^{2})^{\frac{10}{9}}

Add a constant to the solution

= \frac{9}{20} (1 + x^{2})^{\frac{10}{9}} + C

What is the integral of x\sqrt[9]{1+x^2} ?
The integral of x\sqrt[9]{1+x^2} is 9/20 (1+x^2)^{10/9}+C