What is the integral of x^4(x^5+1)^{18} ?

The integral of x^4(x^5+1)^{18} is 1/95 (x^5+1)^{19}+C

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integral of x^4(x^5+1)^{18}

\int x^{4} (x^{5} + 1)^{18} d x

\frac{1}{95} (x^{5} + 1)^{19} + C

Solution steps

\int x^{4} (x^{5} + 1)^{18} d x

Apply u-substitution

= \int \frac{u ^{18}}{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^{18} d u

Apply the Power Rule

= \frac{1}{5} \cdot \frac{u ^{19}}{19}

Substitute back

u = x^{5} + 1

= \frac{1}{5} \cdot \frac{( x ^{5} + 1 ) ^{19}}{19}

Simplify

\frac{1}{5} \cdot \frac{( x ^{5} + 1 ) ^{19}}{19} : \frac{1}{95} (x^{5} + 1)^{19}

= \frac{1}{95} (x^{5} + 1)^{19}

Add a constant to the solution

= \frac{1}{95} (x^{5} + 1)^{19} + C

What is the integral of x^4(x^5+1)^{18} ?
The integral of x^4(x^5+1)^{18} is 1/95 (x^5+1)^{19}+C