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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

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