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

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

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

\int_{- 1}^{1} \frac{tan ( x )}{1 + x ^{4}} d x

0

Solution steps

\int_{- 1}^{1} \frac{tan ( x )}{1 + x ^{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{tan ( x )}{1 + x ^{4}} :

Odd

= 0

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

x = y^{4}, 2 - x, y = 0

\frac{4 x - 6}{2 x ^{2} + x - 2}

\int \frac{8 x ^{2}}{25 - x ^{2}} d x

\frac{d}{d x} (e^{3 y - x^{2}})

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