What is the integral of (x^{11})/(sqrt(x^6-1)) ?

The integral of (x^{11})/(sqrt(x^6-1)) is 1/9 (x^6+2)sqrt(x^6-1)+C

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integral of (x^{11})/(sqrt(x^6-1))

\int \frac{x ^{11}}{x ^{6} - 1} d x

\frac{1}{9} (x^{6} + 2) x^{6} - 1 + C

Solution steps

\int \frac{x ^{11}}{x ^{6} - 1} d x

Apply u-substitution

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

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

Expand

\frac{u + 1}{u} : u + \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int u d u + \int \frac{1}{u} d u)

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{1}{u} d u = 2 u

= \frac{1}{6} (\frac{2}{3} u^{\frac{3}{2}} + 2 u)

Substitute back

u = x^{6} - 1

= \frac{1}{6} (\frac{2}{3} (x^{6} - 1)^{\frac{3}{2}} + 2 x^{6} - 1)

Simplify

\frac{1}{6} (\frac{2}{3} (x^{6} - 1)^{\frac{3}{2}} + 2 x^{6} - 1) : \frac{1}{9} (x^{6} + 2) x^{6} - 1

= \frac{1}{9} (x^{6} + 2) x^{6} - 1

Add a constant to the solution

= \frac{1}{9} (x^{6} + 2) x^{6} - 1 + C

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What is the integral of (x^{11})/(sqrt(x^6-1)) ?
The integral of (x^{11})/(sqrt(x^6-1)) is 1/9 (x^6+2)sqrt(x^6-1)+C