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

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

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

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

0

Solution steps

\int_{- 1}^{1} \frac{2 x}{x ^{2} + 1} 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{2 x}{x ^{2} + 1} :

Odd

= 0

\int_{- 1}^{1} ∣ x ∣ cos (nπ x) d x

\int_{- 1}^{0} x^{3} (x^{2} - 1)^{10} d x

\int_{- 2}^{0} 2 x^{3} + 6 x^{2} + 4 x d x

\int_{0}^{8} (x - \frac{1}{2}) d x

\int_{0}^{1} x^{2} - 3 x + 5 d x

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