What is the integral from-1 to 1 of (5r)/((4+r^2)^2) ?

The integral from-1 to 1 of (5r)/((4+r^2)^2) is 0

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integral from-1 to 1 of (5r)/((4+r^2)^2)

\int_{- 1}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

0

Solution steps

\int_{- 1}^{1} \frac{5 r}{( 4 + r ^{2} ) ^{2}} d r

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

[- a, a]

then

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

Parity of

\frac{5 r}{( 4 + r ^{2} ) ^{2}} :

Odd

= 0

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + x y + y ^{2}})

\int \frac{3 x - 2}{5 x ^{2} - 3 x + 2} d x

55 cos (0.03 x - 1.11) + 86

y^{^{''}} - 2 y^{^{'}} - 3 y = 9

5, x^{2} + 1

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