What is the integral from 0 to 1 of (x^2+1)/(x^4+1) ?

The integral from 0 to 1 of (x^2+1)/(x^4+1) is (8arctan(sqrt(2)+1)+8arctan(sqrt(2)-1))/(8sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of (x^2+1)/(x^4+1)

\int_{0}^{1} \frac{x ^{2} + 1}{x ^{4} + 1} d x

\frac{8 arctan ( 2 + 1 ) + 8 arctan ( 2 - 1 )}{8 2}

Solution steps

\int_{0}^{1} \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_{0}^{1} \frac{x ^{2}}{x ^{4} + 1} d x + \int_{0}^{1} \frac{1}{x ^{4} + 1} d x

\int_{0}^{1} \frac{x ^{2}}{x ^{4} + 1} d x = \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

\int_{0}^{1} \frac{1}{x ^{4} + 1} d x = \frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

= \frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2} + \frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2}

Simplify

\frac{- 2 ln ( 2 + 2 ) + 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2} + \frac{2 ln ( 2 + 2 ) - 2 ln ( 2 - 2 ) + 4 arctan ( 2 + 1 ) + 4 arctan ( 2 - 1 )}{8 2} : \frac{8 arctan ( 2 + 1 ) + 8 arctan ( 2 - 1 )}{8 2}

= \frac{8 arctan ( 2 + 1 ) + 8 arctan ( 2 - 1 )}{8 2}

\int_{- 1}^{1} (∣ x ∣ + x) e^{- ∣ x ∣} d x

\int_{2}^{5} 3 x^{2} d x

\int_{π}^{2 π} (x - 2 sin (x)) d x

\int_{1}^{7} \frac{x + x ^{3}}{x} d x

What is the integral from 0 to 1 of (x^2+1)/(x^4+1) ?
The integral from 0 to 1 of (x^2+1)/(x^4+1) is (8arctan(sqrt(2)+1)+8arctan(sqrt(2)-1))/(8sqrt(2))