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

The integral of (1+x^3)^4 is x+x^4+(6x^7)/7+(2x^{10})/5+(x^{13})/(13)+C

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integral of (1+x^3)^4

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

x + x^{4} + \frac{6 x ^{7}}{7} + \frac{2 x ^{10}}{5} + \frac{x ^{13}}{13} + C

Solution steps

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

Expand

(1 + x^{3})^{4} : 1 + 4 x^{3} + 6 x^{6} + 4 x^{9} + x^{12}

= \int 1 + 4 x^{3} + 6 x^{6} + 4 x^{9} + x^{12} 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 1 d x + \int 4 x^{3} d x + \int 6 x^{6} d x + \int 4 x^{9} d x + \int x^{12} d x

\int 1 d x = x

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

\int 6 x^{6} d x = \frac{6 x ^{7}}{7}

\int 4 x^{9} d x = \frac{2 x ^{10}}{5}

\int x^{12} d x = \frac{x ^{13}}{13}

= x + x^{4} + \frac{6 x ^{7}}{7} + \frac{2 x ^{10}}{5} + \frac{x ^{13}}{13}

Add a constant to the solution

= x + x^{4} + \frac{6 x ^{7}}{7} + \frac{2 x ^{10}}{5} + \frac{x ^{13}}{13} + C

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\int x e^{- x} - x^{2} d x

What is the integral of (1+x^3)^4 ?
The integral of (1+x^3)^4 is x+x^4+(6x^7)/7+(2x^{10})/5+(x^{13})/(13)+C