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

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

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

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

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

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{x ^{4} + 4} d x = - \frac{1}{8} (ln x^{2} + 2 x + 2 - 2 arctan (x + 1)) + \frac{1}{8} (ln x^{2} - 2 x + 2 + 2 arctan (x - 1))

\int \frac{2}{x ^{4} + 4} d x = 2 (\frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1)))

= - \frac{1}{8} (ln x^{2} + 2 x + 2 - 2 arctan (x + 1)) + \frac{1}{8} (ln x^{2} - 2 x + 2 + 2 arctan (x - 1)) + 2 (\frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1)))

Simplify

- \frac{1}{8} (ln x^{2} + 2 x + 2 - 2 arctan (x + 1)) + \frac{1}{8} (ln x^{2} - 2 x + 2 + 2 arctan (x - 1)) + 2 (\frac{1}{16} (ln x^{2} + 2 x + 2 + 2 arctan (x + 1)) + \frac{1}{16} (- ln x^{2} - 2 x + 2 + 2 arctan (x - 1))) : \frac{1}{2} arctan (x + 1) + \frac{1}{2} arctan (x - 1)

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

Add a constant to the solution

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

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

\int \frac{7 x ^{2} + 5 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

f (x) = x^{4 x}

x \to 1 lim (- 6 x sin (3 x^{2} + 1))

f (x) = \frac{4 x}{x + 2}

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