What is the integral of 1/(x^4+x^6) ?

The integral of 1/(x^4+x^6) is 1/x-1/(3x^3)+arctan(x)+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^4+x^6)

\int \frac{1}{x ^{4} + x ^{6}} d x

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

Solution steps

\int \frac{1}{x ^{4} + x ^{6}} d x

Take the partial fraction of

\frac{1}{x ^{4} + x ^{6}} : - \frac{1}{x ^{2}} + \frac{1}{x ^{4}} + \frac{1}{x ^{2} + 1}

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

\int \frac{1}{36 + 9 x ^{2}} d x

\int sec^{2} (x) sin (tan (x)) d x

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

\int e^{2 x - y} d x

What is the integral of 1/(x^4+x^6) ?
The integral of 1/(x^4+x^6) is 1/x-1/(3x^3)+arctan(x)+C