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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

Add a constant to the solution

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

\int \frac{ln ( tan ( x ) )}{sin ( x ) cos ( x )} d x

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

\int (x + e^{x}) x d x

\int \frac{1}{4 + 25 x ^{2}} d x

\int \frac{x cos ( x )}{sin ^{3} ( x )} d x

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