What is the integral of (x^8)/(1+x^{18)} ?

The integral of (x^8)/(1+x^{18)} is 1/9 arctan(x^9)+C

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integral of (x^8)/(1+x^{18)}

\int \frac{x ^{8}}{1 + x ^{18}} d x

\frac{1}{9} arctan (x^{9}) + C

Solution steps

\int \frac{x ^{8}}{1 + x ^{18}} d x

Apply u-substitution

: \int \frac{1}{9 ( 1 + u ^{2} )} d u

= \int \frac{1}{9 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= \frac{1}{9} arctan (u)

Substitute back

u = x^{9}

= \frac{1}{9} arctan (x^{9})

Add a constant to the solution

= \frac{1}{9} arctan (x^{9}) + C

\frac{d}{d x} (2 x e^{x^{2} - 1})

f (30) = \frac{850}{d}

\frac{d}{d x} (\frac{5}{x + 7})

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

\frac{d y}{d x} = \frac{cos ( 18 x )}{e ^{18 y}}

What is the integral of (x^8)/(1+x^{18)} ?
The integral of (x^8)/(1+x^{18)} is 1/9 arctan(x^9)+C