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

The integral of x^2(2x^3-5)^3 is 1/24 (2x^3-5)^4+C

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

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

\frac{1}{24} (2 x^{3} - 5)^{4} + C

Solution steps

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

Apply u-substitution

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{3} - 5 ) ^{4}}{4}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{( 2 x ^{3} - 5 ) ^{4}}{4} : \frac{1}{24} (2 x^{3} - 5)^{4}

= \frac{1}{24} (2 x^{3} - 5)^{4}

Add a constant to the solution

= \frac{1}{24} (2 x^{3} - 5)^{4} + C

\int_{0}^{\frac{π}{4}} sec (t) d t

in v erse l a pl a ce \frac{1}{( s ^{4} + 1 )}

a re a y = 2 (x + 1), y = 3 (x + 1), x = 6

d er i v a t i v e 6 e^{x} + x

in v erse l a pl a ce \frac{s}{s ^{2} - π}

What is the integral of x^2(2x^3-5)^3 ?
The integral of x^2(2x^3-5)^3 is 1/24 (2x^3-5)^4+C