What is the integral of x^2sqrt(ax^3-b) ?

The integral of x^2sqrt(ax^3-b) is (2(ax^3-b)^{3/2})/(9a)+C

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integral of x^2sqrt(ax^3-b)

\int x^{2} a x^{3} - b d x

\frac{2 ( a x ^{3} - b ) ^{\frac{3}{2}}}{9 a} + C

Solution steps

\int x^{2} a x^{3} - b d x

Apply u-substitution

: \int \frac{u a u ^{\frac{3}{2}} - b}{2 x} d u

= \int \frac{a u - b}{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 a u - b d u

Apply u-substitution

: \int \frac{v}{a} d v

= \frac{1}{3} \cdot \int \frac{v}{a} 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}{a} \cdot \int v d v

Apply the Power Rule

: \frac{2}{3} v^{\frac{3}{2}}

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

Substitute back

= \frac{1}{3} \cdot \frac{1}{a} \cdot \frac{2}{3} (a x^{3} - b)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{a} \cdot \frac{2}{3} (a x^{3} - b)^{\frac{3}{2}} : \frac{2 ( a x ^{3} - b ) ^{\frac{3}{2}}}{9 a}

= \frac{2 ( a x ^{3} - b ) ^{\frac{3}{2}}}{9 a}

Add a constant to the solution

= \frac{2 ( a x ^{3} - b ) ^{\frac{3}{2}}}{9 a} + C

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What is the integral of x^2sqrt(ax^3-b) ?
The integral of x^2sqrt(ax^3-b) is (2(ax^3-b)^{3/2})/(9a)+C