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

The integral of 1/(x^4(1+x^4)) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{1}{x ^{4} ( 1 + x ^{4} )} d x

- \frac{1}{3 x ^{3}} + \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)) + C

Solution steps

\int \frac{1}{x ^{4} ( 1 + x ^{4} )} d x

Take the partial fraction of

\frac{1}{x ^{4} ( 1 + x ^{4} )} : \frac{1}{x ^{4}} + \frac{- x - 2}{2 2 ( x ^{2} + 2 x + 1 )} + \frac{x - 2}{2 2 ( x ^{2} - 2 x + 1 )}

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

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{- x - 2}{2 2 ( x ^{2} + 2 x + 1 )} d x = \frac{1}{4 2} (- ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1))

\int \frac{x - 2}{2 2 ( x ^{2} - 2 x + 1 )} d x = \frac{1}{4 2} (ln 2 x^{2} - 22 x + 2 - 2 arctan (2 x - 1))

= - \frac{1}{3 x ^{3}} + \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))

Add a constant to the solution

= - \frac{1}{3 x ^{3}} + \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)) + C

\int \frac{2 sin ( x )}{cos ( x ) cos ( 2 x )} d x

\int \frac{x ^{3}}{( x ^{2} + a ^{2} ) ^{\frac{3}{2}}} d x

\int x^{2} cos (x^{3} + 4) d x

\int \frac{cos ( x )}{sin ( x ) + 1} d x

\int sec^{3} (p) tan (p) d p