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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4+1)/(x^3+2x)

\int \frac{x ^{4} + 1}{x ^{3} + 2 x} d x

\frac{1}{2} x^{2} + \frac{1}{2} ln ∣ x ∣ - \frac{5}{4} ln x^{2} + 2 + 1 + C

Solution steps

\int \frac{x ^{4} + 1}{x ^{3} + 2 x} d x

Expand

\frac{x ^{4} + 1}{x ^{3} + 2 x} : \frac{x ^{4}}{x ^{3} + 2 x} + \frac{1}{x ^{3} + 2 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 ^{4}}{x ^{3} + 2 x} d x + \int \frac{1}{x ^{3} + 2 x} d x

\int \frac{x ^{4}}{x ^{3} + 2 x} d x = \frac{1}{2} x^{2} + 1 - ln x^{2} + 2

\int \frac{1}{x ^{3} + 2 x} d x = \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln x^{2} + 2

= \frac{1}{2} x^{2} + 1 - ln x^{2} + 2 + \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln x^{2} + 2

Simplify

\frac{1}{2} x^{2} + 1 - ln x^{2} + 2 + \frac{1}{2} ln ∣ x ∣ - \frac{1}{4} ln x^{2} + 2 : \frac{1}{2} x^{2} + \frac{1}{2} ln ∣ x ∣ - \frac{5}{4} ln x^{2} + 2 + 1

= \frac{1}{2} x^{2} + \frac{1}{2} ln ∣ x ∣ - \frac{5}{4} ln x^{2} + 2 + 1

Add a constant to the solution

= \frac{1}{2} x^{2} + \frac{1}{2} ln ∣ x ∣ - \frac{5}{4} ln x^{2} + 2 + 1 + C

\int y cos (x y) + y + 1 d x

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

\int 4 x^{3} - 9 + 2 sin (x) + 7 e^{x} d x

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

\int 4 e^{5 t} sin (3 t) d t

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