What is the integral of ((x^2+1))/((x^4+1)) ?

The integral of ((x^2+1))/((x^4+1)) is 1/(sqrt(2))arctan(sqrt(2)x+1)+1/(sqrt(2))arctan(sqrt(2)x-1)+C

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integral of ((x^2+1))/((x^4+1))

\int \frac{( x ^{2} + 1 )}{( x ^{4} + 1 )} d x

\frac{1}{2} arctan (2 x + 1) + \frac{1}{2} arctan (2 x - 1) + C

Solution steps

\int \frac{x ^{2} + 1}{x ^{4} + 1} d x

Expand

\frac{x ^{2} + 1}{x ^{4} + 1} : \frac{x ^{2}}{x ^{4} + 1} + \frac{1}{x ^{4} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{2}}{x ^{4} + 1} d x + \int \frac{1}{x ^{4} + 1} d x

\int \frac{x ^{2}}{x ^{4} + 1} d x = - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

\int \frac{1}{x ^{4} + 1} d x = \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

= - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) + \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1))

Simplify

- \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)) + \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) + \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 + 2 arctan (2 x + 1)) + \frac{1}{4 2} (- ln 2 x^{2} - 22 x + 2 + 2 arctan (2 x - 1)) : \frac{1}{2} arctan (2 x + 1) + \frac{1}{2} arctan (2 x - 1)

= \frac{1}{2} arctan (2 x + 1) + \frac{1}{2} arctan (2 x - 1)

Add a constant to the solution

= \frac{1}{2} arctan (2 x + 1) + \frac{1}{2} arctan (2 x - 1) + C

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What is the integral of ((x^2+1))/((x^4+1)) ?
The integral of ((x^2+1))/((x^4+1)) is 1/(sqrt(2))arctan(sqrt(2)x+1)+1/(sqrt(2))arctan(sqrt(2)x-1)+C