What is the integral from-1 to x^2 of (1+x^3)^4 ?

The integral from-1 to x^2 of (1+x^3)^4 is 1/13 (x^{26}+1)+x^8+x^2+6/7 (x^{14}+1)+2/5 (x^{20}-1)

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integral from-1 to x^2 of (1+x^3)^4

\int_{- 1}^{x^{2}} (1 + x^{3})^{4} d x

\frac{1}{13} (x^{26} + 1) + x^{8} + x^{2} + \frac{6}{7} (x^{14} + 1) + \frac{2}{5} (x^{20} - 1)

Solution steps

\int_{- 1}^{x^{2}} (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}^{x^{2}} 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}^{x^{2}} 1 d x + \int_{- 1}^{x^{2}} 4 x^{3} d x + \int_{- 1}^{x^{2}} 6 x^{6} d x + \int_{- 1}^{x^{2}} 4 x^{9} d x + \int_{- 1}^{x^{2}} x^{12} d x

\int_{- 1}^{x^{2}} 1 d x = x^{2} + 1

\int_{- 1}^{x^{2}} 4 x^{3} d x = x^{8} - 1

\int_{- 1}^{x^{2}} 6 x^{6} d x = \frac{6 ( x ^{14} + 1 )}{7}

\int_{- 1}^{x^{2}} 4 x^{9} d x = \frac{2 ( x ^{20} - 1 )}{5}

\int_{- 1}^{x^{2}} x^{12} d x = \frac{x ^{26} + 1}{13}

= x^{2} + 1 + x^{8} - 1 + \frac{6 ( x ^{14} + 1 )}{7} + \frac{2 ( x ^{20} - 1 )}{5} + \frac{x ^{26} + 1}{13}

Simplify

x^{2} + 1 + x^{8} - 1 + \frac{6 ( x ^{14} + 1 )}{7} + \frac{2 ( x ^{20} - 1 )}{5} + \frac{x ^{26} + 1}{13} : \frac{1}{13} (x^{26} + 1) + x^{8} + x^{2} + \frac{6}{7} (x^{14} + 1) + \frac{2}{5} (x^{20} - 1)

= \frac{1}{13} (x^{26} + 1) + x^{8} + x^{2} + \frac{6}{7} (x^{14} + 1) + \frac{2}{5} (x^{20} - 1)

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What is the integral from-1 to x^2 of (1+x^3)^4 ?
The integral from-1 to x^2 of (1+x^3)^4 is 1/13 (x^{26}+1)+x^8+x^2+6/7 (x^{14}+1)+2/5 (x^{20}-1)