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

The integral from-1 to 1 of x/((x^2+1)^2) is 0

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

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

0

Solution steps

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

\frac{x}{( x ^{2} + 1 ) ^{2}} :

Odd

= 0

f (x) = tan (4 x)

4

- 4 cos (t)

n = 0 \sum \infty \frac{3 n}{4 n - 1}

y = arctan (ln (2) x^{5})

What is the integral from-1 to 1 of x/((x^2+1)^2) ?
The integral from-1 to 1 of x/((x^2+1)^2) is 0