What is the integral of 2x^2(-x^3+3)^5 ?

The integral of 2x^2(-x^3+3)^5 is -1/9 (-x^3+3)^6+C

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integral of 2x^2(-x^3+3)^5

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

- \frac{1}{9} (- x^{3} + 3)^{6} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ( - u ^{\frac{3}{2}} + 3 ) ^{5}}{2 x} d u

= 2 \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

= 2 \cdot \frac{1}{3} \cdot \int (- u + 3)^{5} d u

Apply u-substitution

: \int - v^{5} d v

= 2 \cdot \frac{1}{3} \cdot \int - v^{5} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} (- \int v^{5} d v)

Apply the Power Rule

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

= 2 \cdot \frac{1}{3} (- \frac{v ^{6}}{6})

Substitute back

= 2 \cdot \frac{1}{3} (- \frac{( - x ^{3} + 3 ) ^{6}}{6})

Simplify

2 \cdot \frac{1}{3} (- \frac{( - x ^{3} + 3 ) ^{6}}{6}) : - \frac{1}{9} (- x^{3} + 3)^{6}

= - \frac{1}{9} (- x^{3} + 3)^{6}

Add a constant to the solution

= - \frac{1}{9} (- x^{3} + 3)^{6} + C

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What is the integral of 2x^2(-x^3+3)^5 ?
The integral of 2x^2(-x^3+3)^5 is -1/9 (-x^3+3)^6+C