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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

2 e^{x} cos (x)

\frac{\partial}{\partial x} (ln (x^{3}))

\int \frac{1}{( 1 - y )} d y

\int (\frac{3}{x}) d x

\frac{\partial}{\partial z} (5 sin^{2} (z))

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