What is the integral of x^7sqrt(x^4+4) ?

The integral of x^7sqrt(x^4+4) is 1/2 (1/5 (x^4+4)^{5/2}-4/3 (x^4+4)^{3/2})+C

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integral of x^7sqrt(x^4+4)

\int x^{7} x^{4} + 4 d x

\frac{1}{2} (\frac{1}{5} (x^{4} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{4} + 4)^{\frac{3}{2}}) + C

Solution steps

\int x^{7} x^{4} + 4 d x

Apply u-substitution

: \int \frac{( u ^{2} - 4 ) u ^{2}}{2} d u

= \int \frac{( u ^{2} - 4 ) u ^{2}}{2} d u

Take the constant out:

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

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

Expand

(u^{2} - 4) u^{2} : u^{4} - 4 u^{2}

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

= \frac{1}{2} (\frac{u ^{5}}{5} - \frac{4 u ^{3}}{3})

Substitute back

u = x^{4} + 4

= \frac{1}{2} (\frac{( x ^{4} + 4 ) ^{5}}{5} - \frac{4 ( x ^{4} + 4 ) ^{3}}{3})

Simplify

\frac{1}{2} (\frac{( x ^{4} + 4 ) ^{5}}{5} - \frac{4 ( x ^{4} + 4 ) ^{3}}{3}) : \frac{1}{2} (\frac{1}{5} (x^{4} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{4} + 4)^{\frac{3}{2}})

= \frac{1}{2} (\frac{1}{5} (x^{4} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{4} + 4)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{2} (\frac{1}{5} (x^{4} + 4)^{\frac{5}{2}} - \frac{4}{3} (x^{4} + 4)^{\frac{3}{2}}) + C

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What is the integral of x^7sqrt(x^4+4) ?
The integral of x^7sqrt(x^4+4) is 1/2 (1/5 (x^4+4)^{5/2}-4/3 (x^4+4)^{3/2})+C