What is the integral of x^2(a+bx^3)^2 ?

The integral of x^2(a+bx^3)^2 is ((a+bx^3)^3)/(9b)+C

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integral of x^2(a+bx^3)^2

\int x^{2} (a + b x^{3})^{2} d x

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

Solution steps

\int x^{2} (a + b x^{3})^{2} d x

Apply u-substitution

= \int \frac{( a + b u ) ^{2}}{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 + b u)^{2} d u

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of x^2(a+bx^3)^2 ?
The integral of x^2(a+bx^3)^2 is ((a+bx^3)^3)/(9b)+C