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

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

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

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

0

Solution steps

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

Odd

= 0

\int_{- 6}^{7} ∣ x ∣ d x

\int_{0}^{1} \frac{2 x}{( x ^{2} + 5 ) ^{3}} d x

\int_{- 2}^{3} ∣ 4 x (x - 1) ∣ d x

\int_{0}^{2} 2 x e^{- x} d x

\int_{0}^{1} e^{y^{2}} d y

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