What is the integral of x^3(x^4+3)^2 ?

The integral of x^3(x^4+3)^2 is (x^{12}}{12}+\frac{3x^8)/4+(9x^4)/4+C

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integral of x^3(x^4+3)^2

\int x^{3} (x^{4} + 3)^{2} d x

\frac{x ^{12}}{12} + \frac{3 x ^{8}}{4} + \frac{9 x ^{4}}{4} + C

Solution steps

\int x^{3} (x^{4} + 3)^{2} d x

Expand

x^{3} (x^{4} + 3)^{2} : x^{11} + 6 x^{7} + 9 x^{3}

= \int x^{11} + 6 x^{7} + 9 x^{3} 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 x^{11} d x + \int 6 x^{7} d x + \int 9 x^{3} d x

\int x^{11} d x = \frac{x ^{12}}{12}

\int 6 x^{7} d x = \frac{3 x ^{8}}{4}

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

= \frac{x ^{12}}{12} + \frac{3 x ^{8}}{4} + \frac{9 x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{12}}{12} + \frac{3 x ^{8}}{4} + \frac{9 x ^{4}}{4} + C

y = x^{5} - 4 x^{3} + 2 x - 3

\frac{8 x}{1 - x}

- 2 t + 8

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

\frac{d}{d x} (- \frac{2 x}{( 9 + x ^{2} ) ^{2}})

What is the integral of x^3(x^4+3)^2 ?
The integral of x^3(x^4+3)^2 is (x^{12}}{12}+\frac{3x^8)/4+(9x^4)/4+C