What is the integral of (x^3-3)^56x^2 ?

The integral of (x^3-3)^56x^2 is 1/3 (x^3-3)^6+C

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integral of (x^3-3)^56x^2

\int (x^{3} - 3)^{5} 6 x^{2} d x

\frac{1}{3} (x^{3} - 3)^{6} + C

Solution steps

\int (x^{3} - 3)^{5} \cdot 6 x^{2} d x

Take the constant out:

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

= 6 \cdot \int (x^{3} - 3)^{5} x^{2} d x

Apply u-substitution

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

= 6 \cdot \int \frac{( u - 3 ) ^{5}}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int (u - 3)^{5} d u

Apply u-substitution

: \int v^{5} d v

= 6 \cdot \frac{1}{3} \cdot \int v^{5} d v

Apply the Power Rule

: \frac{v ^{6}}{6}

= 6 \cdot \frac{1}{3} \cdot \frac{v ^{6}}{6}

Substitute back

= 6 \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 3 ) ^{6}}{6}

Simplify

6 \cdot \frac{1}{3} \cdot \frac{( x ^{3} - 3 ) ^{6}}{6} : \frac{1}{3} (x^{3} - 3)^{6}

= \frac{1}{3} (x^{3} - 3)^{6}

Add a constant to the solution

= \frac{1}{3} (x^{3} - 3)^{6} + C

\int csc (x + 6) d x

4 - x^{2}

\int \frac{- 2}{1 - x ^{2}} d x

\int \frac{1}{tan ( t )} d t

What is the integral of (x^3-3)^56x^2 ?
The integral of (x^3-3)^56x^2 is 1/3 (x^3-3)^6+C