What is the integral of ((x^5+1))/(sqrt(x^6+6x)) ?

The integral of ((x^5+1))/(sqrt(x^6+6x)) is 1/3 sqrt(x^6+6x)+C

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integral of ((x^5+1))/(sqrt(x^6+6x))

\int \frac{( x ^{5} + 1 )}{x ^{6} + 6 x} d x

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

Solution steps

\int \frac{x ^{5} + 1}{x ^{6} + 6 x} 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 = x^{6} + 6 x

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

Simplify

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

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

Add a constant to the solution

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

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