What is the integral from-1 to 1 of (x^5-6x^9+(sin(x))/((1+x^4)^2)) ?

The integral from-1 to 1 of (x^5-6x^9+(sin(x))/((1+x^4)^2)) is 0

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integral from-1 to 1 of (x^5-6x^9+(sin(x))/((1+x^4)^2))

\int_{- 1}^{1} (x^{5} - 6 x^{9} + \frac{sin ( x )}{( 1 + x ^{4} ) ^{2}}) d x

0

Solution steps

\int_{- 1}^{1} x^{5} - 6 x^{9} + \frac{sin ( x )}{( 1 + x ^{4} ) ^{2}} 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^{5} - 6 x^{9} + \frac{sin ( x )}{( 1 + x ^{4} ) ^{2}} :

Odd

= 0

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What is the integral from-1 to 1 of (x^5-6x^9+(sin(x))/((1+x^4)^2)) ?
The integral from-1 to 1 of (x^5-6x^9+(sin(x))/((1+x^4)^2)) is 0