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

The integral from-1 to 1 of x^3(1+x^4)^3 is 0

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

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

0

Solution steps

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

If f(x) is an odd function and continuous at:

[- a, a]

then

\int_{- a}^{a} f (x) d x = 0

Parity of

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

Odd

= 0

f (x) = 8 + 5 x^{2} - 2 x^{3}, (1, 11)

f (x) = \frac{4}{x}, (4, 1)

\int_{0}^{l n (3)} 2 (3 - x e^{x}) d x

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

\int x 3^{x^{2}} d x

What is the integral from-1 to 1 of x^3(1+x^4)^3 ?
The integral from-1 to 1 of x^3(1+x^4)^3 is 0