What is the integral of x^2(x^3+1)^9 ?

The integral of x^2(x^3+1)^9 is 1/30 (x^3+1)^{10}+C

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integral of x^2(x^3+1)^9

\int x^{2} (x^{3} + 1)^{9} d x

\frac{1}{30} (x^{3} + 1)^{10} + C

Solution steps

\int x^{2} (x^{3} + 1)^{9} d x

Apply u-substitution

= \int \frac{( u + 1 ) ^{9}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int (u + 1)^{9} d u

Apply u-substitution

= \frac{1}{3} \cdot \int v^{9} d v

Apply the Power Rule

= \frac{1}{3} \cdot \frac{v ^{10}}{10}

Substitute back

= \frac{1}{3} \cdot \frac{( x ^{3} + 1 ) ^{10}}{10}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 1 ) ^{10}}{10} : \frac{1}{30} (x^{3} + 1)^{10}

= \frac{1}{30} (x^{3} + 1)^{10}

Add a constant to the solution

= \frac{1}{30} (x^{3} + 1)^{10} + C

What is the integral of x^2(x^3+1)^9 ?
The integral of x^2(x^3+1)^9 is 1/30 (x^3+1)^{10}+C