What is the integral of ((a+bx^5))/(sqrt(6ax+bx^6)) ?

The integral of ((a+bx^5))/(sqrt(6ax+bx^6)) is 1/3 sqrt(6ax+bx^6)+C

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integral of ((a+bx^5))/(sqrt(6ax+bx^6))

\int \frac{( a + b x ^{5} )}{6 a x + b x ^{6}} d x

\frac{1}{3} 6 a x + b x^{6} + C

Solution steps

\int \frac{a + b x ^{5}}{6 a x + b x ^{6}} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 6 a x + b x^{6}

= \frac{1}{6} \cdot 2 (6 a x + b x^{6})^{\frac{1}{2}}

Simplify

\frac{1}{6} \cdot 2 (6 a x + b x^{6})^{\frac{1}{2}} : \frac{1}{3} 6 a x + b x^{6}

= \frac{1}{3} 6 a x + b x^{6}

Add a constant to the solution

= \frac{1}{3} 6 a x + b x^{6} + C

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What is the integral of ((a+bx^5))/(sqrt(6ax+bx^6)) ?
The integral of ((a+bx^5))/(sqrt(6ax+bx^6)) is 1/3 sqrt(6ax+bx^6)+C