What is the integral of x^3(1+x^4)^3 ?

The integral of x^3(1+x^4)^3 is (x^4)/4+(3x^8)/8+(x^{12})/4+(x^{16})/(16)+C

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integral of x^3(1+x^4)^3

integral

x^{3} (1 + x^{4})^{3}

\frac{x ^{4}}{4} + \frac{3 x ^{8}}{8} + \frac{x ^{12}}{4} + \frac{x ^{16}}{16} + C

Solution steps

\int x^{3} (1 + x^{4})^{3} d x

Expand

x^{3} (1 + x^{4})^{3} : x^{3} + 3 x^{7} + 3 x^{11} + x^{15}

= \int x^{3} + 3 x^{7} + 3 x^{11} + x^{15} 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^{3} d x + \int 3 x^{7} d x + \int 3 x^{11} d x + \int x^{15} d x

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

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

\int 3 x^{11} d x = \frac{x ^{12}}{4}

\int x^{15} d x = \frac{x ^{16}}{16}

= \frac{x ^{4}}{4} + \frac{3 x ^{8}}{8} + \frac{x ^{12}}{4} + \frac{x ^{16}}{16}

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{3 x ^{8}}{8} + \frac{x ^{12}}{4} + \frac{x ^{16}}{16} + C

\frac{( 1 - x )}{x ^{2}}

\frac{x}{( 2 - x )}

\frac{e ^{- 1}}{x}

(x - 1)^{2}

(e^{x} + e^{- x})^{4}

What is the integral of x^3(1+x^4)^3 ?
The integral of x^3(1+x^4)^3 is (x^4)/4+(3x^8)/8+(x^{12})/4+(x^{16})/(16)+C