What is the integral of x^9(5+x^5)^{99} ?

The integral of x^9(5+x^5)^{99} is 1/5 (((5+x^5)^{101})/(101)-((5+x^5)^{100})/(20))+C

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integral of x^9(5+x^5)^{99}

\int x^{9} (5 + x^{5})^{99} d x

\frac{1}{5} (\frac{( 5 + x ^{5} ) ^{101}}{101} - \frac{( 5 + x ^{5} ) ^{100}}{20}) + C

Solution steps

\int x^{9} (5 + x^{5})^{99} d x

Apply u-substitution

: \int \frac{( u - 5 ) u ^{99}}{5} d u

= \int \frac{( u - 5 ) u ^{99}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int (u - 5) u^{99} d u

Expand

(u - 5) u^{99} : u^{100} - 5 u^{99}

= \frac{1}{5} \cdot \int u^{100} - 5 u^{99} 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}{5} (\int u^{100} d u - \int 5 u^{99} d u)

\int u^{100} d u = \frac{u ^{101}}{101}

\int 5 u^{99} d u = \frac{u ^{100}}{20}

= \frac{1}{5} (\frac{u ^{101}}{101} - \frac{u ^{100}}{20})

Substitute back

u = 5 + x^{5}

= \frac{1}{5} (\frac{( 5 + x ^{5} ) ^{101}}{101} - \frac{( 5 + x ^{5} ) ^{100}}{20})

Add a constant to the solution

= \frac{1}{5} (\frac{( 5 + x ^{5} ) ^{101}}{101} - \frac{( 5 + x ^{5} ) ^{100}}{20}) + C

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What is the integral of x^9(5+x^5)^{99} ?
The integral of x^9(5+x^5)^{99} is 1/5 (((5+x^5)^{101})/(101)-((5+x^5)^{100})/(20))+C