What is the integral of (x^{11})/(sqrt(2x^6+2)) ?

The integral of (x^{11})/(sqrt(2x^6+2)) is 1/(3sqrt(2))(1/3 (x^6+1)^{3/2}-sqrt(x^6+1))+C

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integral of (x^{11})/(sqrt(2x^6+2))

\int \frac{x ^{11}}{2 x ^{6} + 2} d x

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

Solution steps

\int \frac{x ^{11}}{2 x ^{6} + 2} d x

\frac{x ^{11}}{2 x ^{6} + 2} = \frac{1}{2} \frac{x ^{11}}{x ^{6} + 1}

= \int \frac{1}{2} \cdot \frac{x ^{11}}{x ^{6} + 1} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x ^{11}}{x ^{6} + 1} d x

Apply u-substitution

: \int \frac{u ^{2} - 1}{3} d u

= \frac{1}{2} \cdot \int \frac{u ^{2} - 1}{3} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int u^{2} - 1 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}{2} \cdot \frac{1}{3} (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= \frac{1}{2} \cdot \frac{1}{3} (\frac{u ^{3}}{3} - u)

Substitute back

u = x^{6} + 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + y^{^{'}} - 6 y = 2 x

t^{2} y^{^{'}} + t y = y

n = 1 \sum \infty \frac{3}{n ^{2} + 3 n}

\frac{d}{d x} (- \frac{A}{2 x ^{\frac{3}{2}}})

x \to 4 lim (\frac{x ^{2} - 1}{x - 4})

What is the integral of (x^{11})/(sqrt(2x^6+2)) ?
The integral of (x^{11})/(sqrt(2x^6+2)) is 1/(3sqrt(2))(1/3 (x^6+1)^{3/2}-sqrt(x^6+1))+C