What is the integral of (x^8+1)/(sqrt(x^9+9x)) ?

The integral of (x^8+1)/(sqrt(x^9+9x)) is 2/9 sqrt(x^9+9x)+C

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integral of (x^8+1)/(sqrt(x^9+9x))

\int \frac{x ^{8} + 1}{x ^{9} + 9 x} d x

\frac{2}{9} x^{9} + 9 x + C

Solution steps

\int \frac{x ^{8} + 1}{x ^{9} + 9 x} d x

Apply u-substitution

: \int \frac{1}{9 u} d u

= \int \frac{1}{9 u} 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}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = x^{9} + 9 x

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

Simplify

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

= \frac{2}{9} x^{9} + 9 x

Add a constant to the solution

= \frac{2}{9} x^{9} + 9 x + C

4^{t}

\frac{d}{d x} (\frac{4}{x + 6})

f (x) = - 4 tan (x) - 1, at x = - \frac{π}{4}

\frac{\partial}{\partial x} (2 y^{6} + 4 y^{8} x^{3})

What is the integral of (x^8+1)/(sqrt(x^9+9x)) ?
The integral of (x^8+1)/(sqrt(x^9+9x)) is 2/9 sqrt(x^9+9x)+C