How calculate integral (down -1, up 1) 1/(e^x+1)(x^2+1)dx ?

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You need to use the following property of definite integrals with symmetrical limits of integration, such that:

`int_(-a)^a f(x)dx = int_0^a(f(x) + f(-x))dx`

You need to evaluate `f(x) + f(-x)` , such that:

`f(x) + f(-x) = 1/((e^x+1)(x^2+1)) + 1/((e^(-x)+1)((-x)^2+1))`

`f(x) + f(-x) = 1/((e^x+1)(x^2+1)) + 1/((1/e^x + 1)(x^2 + 1))`

`f(x) + f(-x) = 1/((e^x+1)(x^2+1)) +e^x/((e^x+1)(x^2+1))`

`f(x) + f(-x) = (e^x + 1)/((e^x+1)(x^2+1))`

Reducing duplicate factors, yields:

`f(x) + f(-x) = 1/(x^2 + 1)`

Hence, you may evaluate the given definite integral, using the indicated property, such that:

`int_(-1)^1 1/((e^x+1)(x^2+1)) dx = int_0^1 1/(x^2 + 1)dx`

`int_0^1 1/(x^2 + 1)dx = tan^(-1) x|_0^1`

`int_0^1 1/(x^2 + 1)dx = tan^(-1) 1 - tan^(-1) 0`

`int_0^1 1/(x^2 + 1)dx = pi/4`

**Hence, evaluating the given definite integral, under the given conditions, yields **`int_(-1)^1 1/((e^x+1)(x^2+1)) dx = pi/4.`

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