What is the integral of (x^7)/(sqrt(x^4-2)) ?

The integral of (x^7)/(sqrt(x^4-2)) is 1/6 (x^4+4)sqrt(x^4-2)+C

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integral of (x^7)/(sqrt(x^4-2))

\int \frac{x ^{7}}{x ^{4} - 2} d x

\frac{1}{6} (x^{4} + 4) x^{4} - 2 + C

Solution steps

\int \frac{x ^{7}}{x ^{4} - 2} d x

Apply u-substitution

: \int \frac{u + 2}{4 u} d u

= \int \frac{u + 2}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u + 2}{u} d u

Expand

\frac{u + 2}{u} : u + \frac{2}{u}

= \frac{1}{4} \cdot \int u + \frac{2}{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}{4} (\int u d u + \int \frac{2}{u} d u)

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

\int \frac{2}{u} d u = 4 u

= \frac{1}{4} (\frac{2}{3} u^{\frac{3}{2}} + 4 u)

Substitute back

u = x^{4} - 2

= \frac{1}{4} (\frac{2}{3} (x^{4} - 2)^{\frac{3}{2}} + 4 x^{4} - 2)

Simplify

\frac{1}{4} (\frac{2}{3} (x^{4} - 2)^{\frac{3}{2}} + 4 x^{4} - 2) : \frac{1}{6} (x^{4} + 4) x^{4} - 2

= \frac{1}{6} (x^{4} + 4) x^{4} - 2

Add a constant to the solution

= \frac{1}{6} (x^{4} + 4) x^{4} - 2 + C

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What is the integral of (x^7)/(sqrt(x^4-2)) ?
The integral of (x^7)/(sqrt(x^4-2)) is 1/6 (x^4+4)sqrt(x^4-2)+C