What is the integral of (x^5+3x)/(x^4+1) ?

The integral of (x^5+3x)/(x^4+1) is (x^2)/2+arctan(x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^5+3x)/(x^4+1)

\int \frac{x ^{5} + 3 x}{x ^{4} + 1} d x

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

Solution steps

\int \frac{x ^{5} + 3 x}{x ^{4} + 1} d x

Expand

\frac{x ^{5} + 3 x}{x ^{4} + 1} : \frac{x ^{5}}{x ^{4} + 1} + \frac{3 x}{x ^{4} + 1}

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 ^{5}}{x ^{4} + 1} d x + \int \frac{3 x}{x ^{4} + 1} d x

\int \frac{x ^{5}}{x ^{4} + 1} d x = \frac{x ^{2}}{2} - \frac{1}{2} arctan (x^{2})

\int \frac{3 x}{x ^{4} + 1} d x = \frac{3}{2} arctan (x^{2})

= \frac{x ^{2}}{2} - \frac{1}{2} arctan (x^{2}) + \frac{3}{2} arctan (x^{2})

Add similar elements:

- \frac{1}{2} arctan (x^{2}) + \frac{3}{2} arctan (x^{2}) = arctan (x^{2})

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

Add a constant to the solution

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

\int \frac{3 x}{( x + 5 ) ( 2 x - 1 )} d x

\int (- \frac{csc ( x ) cot ( x )}{6}) d x

\int \frac{x ^{2} + 7}{x} d x

\int x \cdot 1 0^{x} d x

\int e^{2 t} arctan (e^{t}) d t

What is the integral of (x^5+3x)/(x^4+1) ?
The integral of (x^5+3x)/(x^4+1) is (x^2)/2+arctan(x^2)+C