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

The integral of x(a+bx^3)^2 is (a^2x^2)/2+(2abx^5)/5+(b^2x^8)/8+C

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

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

\frac{a ^{2} x ^{2}}{2} + \frac{2 ab x ^{5}}{5} + \frac{b ^{2} x ^{8}}{8} + C

Solution steps

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

Expand

x (a + b x^{3})^{2} : a^{2} x + 2 ab x^{4} + b^{2} x^{7}

= \int a^{2} x + 2 ab x^{4} + b^{2} x^{7} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int a^{2} x d x + \int 2 ab x^{4} d x + \int b^{2} x^{7} d x

\int a^{2} x d x = \frac{a ^{2} x ^{2}}{2}

\int 2 ab x^{4} d x = \frac{2 ab x ^{5}}{5}

\int b^{2} x^{7} d x = \frac{b ^{2} x ^{8}}{8}

= \frac{a ^{2} x ^{2}}{2} + \frac{2 ab x ^{5}}{5} + \frac{b ^{2} x ^{8}}{8}

Add a constant to the solution

= \frac{a ^{2} x ^{2}}{2} + \frac{2 ab x ^{5}}{5} + \frac{b ^{2} x ^{8}}{8} + C

\int \frac{1}{( x ^{4} - 1 ) x} d x

\int \frac{1}{2 x ^{2} - x ^{3}} d x

\int \frac{1}{( x ^{4} - x ^{2} )} d x

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

\int 5^{x^{2}} \cdot x d x

What is the integral of x(a+bx^3)^2 ?
The integral of x(a+bx^3)^2 is (a^2x^2)/2+(2abx^5)/5+(b^2x^8)/8+C