What is the integral of (x^3-5)^{10}x^2 ?

The integral of (x^3-5)^{10}x^2 is 1/33 (x^3-5)^{11}+C

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integral of (x^3-5)^{10}x^2

\int (x^{3} - 5)^{10} x^{2} d x

\frac{1}{33} (x^{3} - 5)^{11} + C

Solution steps

\int (x^{3} - 5)^{10} x^{2} d x

Apply u-substitution

: \int \frac{( u - 5 ) ^{10}}{3} d u

= \int \frac{( u - 5 ) ^{10}}{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 - 5)^{10} d u

Apply u-substitution

: \int v^{10} d v

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

Apply the Power Rule

: \frac{v ^{11}}{11}

= \frac{1}{3} \cdot \frac{v ^{11}}{11}

Substitute back

= \frac{1}{3} \cdot \frac{( x ^{3} - 5 ) ^{11}}{11}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} - 5 ) ^{11}}{11} : \frac{1}{33} (x^{3} - 5)^{11}

= \frac{1}{33} (x^{3} - 5)^{11}

Add a constant to the solution

= \frac{1}{33} (x^{3} - 5)^{11} + C

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What is the integral of (x^3-5)^{10}x^2 ?
The integral of (x^3-5)^{10}x^2 is 1/33 (x^3-5)^{11}+C