Integration

$\large \int \dfrac1{1+x^4} \, dx = \, ?$

#Calculus

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Comments

A solution without complex numbers,
$I = \displaystyle \int \dfrac{1}{1+x^{4}}dx = \dfrac{1}{2} \int \dfrac{1+x^{2}}{1+x^{4}}dx - \dfrac{x^{2}-1}{1+x^{4}}dx$

$\therefore I = \dfrac{1}{2} \displaystyle \int \dfrac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2}dx - \dfrac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}-2}dx$

In the first integral, substitute $x - \dfrac{1}{x} = u$ and in the second integral substitute $x + \dfrac{1}{x} = v$

$\therefore I = \dfrac{1}{2} \displaystyle \int \dfrac{du}{u^{2}+2} - \int \dfrac{dv}{v^{2}-2}$
I think you can continue after this.

A Former Brilliant Member - 5 years, 1 month ago

thanks

Kalpa Roy - 5 years, 1 month ago

can anyone please help me out with this integration

Kalpa Roy - 5 years, 2 months ago

First application of integration of f(x)/g(x). Then application of integration of f (x)+g(x) In denominator

arpan manchanda - 5 years, 2 months ago

Hint: $1+x^4 = (x^2 + i)(x^2- i )$ for $i = \sqrt{-1}$ , apply partial fractions.

Pi Han Goh - 5 years, 2 months ago

thanx

Kalpa Roy - 5 years, 2 months ago

Se puede factorar cimplentando trinomio cuadrado perfecto (x^2+1-raiz(2)x)(x^2+1+raiz(2)x)

Carlos Suarez - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$