What is the integral of x^7sqrt(9-x^4) ?

The integral of x^7sqrt(9-x^4) is -1/10 (x^4+6)(-x^4+9)^{3/2}+C

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integral of x^7sqrt(9-x^4)

\int x^{7} 9 - x^{4} d x

- \frac{1}{10} (x^{4} + 6) (- x^{4} + 9)^{\frac{3}{2}} + C

Solution steps

\int x^{7} 9 - x^{4} d x

Apply u-substitution

: \int - \frac{( - u + 9 ) u}{4} d u

= \int - \frac{( - u + 9 ) u}{4} 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 (- u + 9) u d u

Expand

(- u + 9) u : - u^{\frac{3}{2}} + 9 u

= - \frac{1}{4} \cdot \int - u^{\frac{3}{2}} + 9 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^{\frac{3}{2}} d u + \int 9 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= - \frac{1}{4} (- \frac{2}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 9 - x^{4}

= - \frac{1}{4} (- \frac{2}{5} (9 - x^{4})^{\frac{5}{2}} + 6 (9 - x^{4})^{\frac{3}{2}})

Simplify

- \frac{1}{4} (- \frac{2}{5} (9 - x^{4})^{\frac{5}{2}} + 6 (9 - x^{4})^{\frac{3}{2}}) : - \frac{1}{10} (x^{4} + 6) (- x^{4} + 9)^{\frac{3}{2}}

= - \frac{1}{10} (x^{4} + 6) (- x^{4} + 9)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{10} (x^{4} + 6) (- x^{4} + 9)^{\frac{3}{2}} + C

\int (3 - 2 x - x^{2})^{3} d x

\int (\frac{cos ^{3} ( x ) + 1}{cos ^{2} ( x )}) d x

\int \frac{x ^{3}}{9 + x ^{4}} d x

\int \frac{10}{x ^{3} + x ^{2}} d x

\int \frac{1}{x ^{3} ( x + 2 ) ^{3}} d x

What is the integral of x^7sqrt(9-x^4) ?
The integral of x^7sqrt(9-x^4) is -1/10 (x^4+6)(-x^4+9)^{3/2}+C