What is the integral of (a+bx^8)/(sqrt(9ax+bx^9)) ?

The integral of (a+bx^8)/(sqrt(9ax+bx^9)) is 2/9 sqrt(9ax+bx^9)+C

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integral of (a+bx^8)/(sqrt(9ax+bx^9))

\int \frac{a + b x ^{8}}{9 a x + b x ^{9}} d x

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

Solution steps

\int \frac{a + b x ^{8}}{9 a x + b x ^{9}} 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 = 9 a x + b x^{9}

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

Simplify

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

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

Add a constant to the solution

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

x \to \infty lim (b^{x})

cos (t - \frac{π}{6})

\int - cos (x) + sin (x) d x

\frac{d}{d x} (\frac{2}{3} x)

What is the integral of (a+bx^8)/(sqrt(9ax+bx^9)) ?
The integral of (a+bx^8)/(sqrt(9ax+bx^9)) is 2/9 sqrt(9ax+bx^9)+C