What is the integral of ((a+bx^9))/(sqrt(10ax+bx^{10))} ?

The integral of ((a+bx^9))/(sqrt(10ax+bx^{10))} is 1/5 sqrt(10ax+bx^{10)}+C

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integral of ((a+bx^9))/(sqrt(10ax+bx^{10))}

\int \frac{( a + b x ^{9} )}{10 a x + b x ^{10}} d x

\frac{1}{5} 10 a x + b x^{10} + C

Solution steps

\int \frac{a + b x ^{9}}{10 a x + b x ^{10}} d x

Apply u-substitution

: \int \frac{1}{10 u} d u

= \int \frac{1}{10 u} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

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

Substitute back

u = 10 a x + b x^{10}

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

Simplify

\frac{1}{10} \cdot 2 (10 a x + b x^{10})^{\frac{1}{2}} : \frac{1}{5} 10 a x + b x^{10}

= \frac{1}{5} 10 a x + b x^{10}

Add a constant to the solution

= \frac{1}{5} 10 a x + b x^{10} + C

\int \frac{x}{( x ^{2} + 1 ) ( x ^{2} + 4 x + 4 )} d x

x \to 0 lim (8 x csc (5 x))

11 (t + 1) \frac{d y}{d t} - 8 y = 24 t

y^{^{'}} + e^{x} y = e^{x}

\frac{d}{d t} (t - \frac{1}{t})

What is the integral of ((a+bx^9))/(sqrt(10ax+bx^{10))} ?
The integral of ((a+bx^9))/(sqrt(10ax+bx^{10))} is 1/5 sqrt(10ax+bx^{10)}+C