What is the integral from-1 to 1 of (8x)/(x^2+4) ?

The integral from-1 to 1 of (8x)/(x^2+4) is 0

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integral from-1 to 1 of (8x)/(x^2+4)

\int_{- 1}^{1} \frac{8 x}{x ^{2} + 4} d x

0

Solution steps

\int_{- 1}^{1} \frac{8 x}{x ^{2} + 4} 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{8 x}{x ^{2} + 4} :

Odd

= 0

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\int_{0}^{\frac{π}{4}} tan (x^{2}) d x

\int_{- 1}^{1} (u^{7} + 1) (3 u^{2} + 1) d u

\int_{0}^{4} x \cdot (3 - 5 x) d x

What is the integral from-1 to 1 of (8x)/(x^2+4) ?
The integral from-1 to 1 of (8x)/(x^2+4) is 0