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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln x^{2} + 3 + ln ∣ x ∣ + C

Solution steps

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

Simplify

\frac{2 x ^{3} + 3 x}{x ^{4} + 3 x ^{2}} : \frac{2 x ^{2} + 3}{x ( x ^{2} + 3 )}

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

Expand

\frac{2 x ^{2} + 3}{x ( x ^{2} + 3 )} : \frac{2 x ^{2}}{x ( x ^{2} + 3 )} + \frac{3}{x ( x ^{2} + 3 )}

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

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

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

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

Simplify

ln x^{2} + 3 + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 3 : \frac{1}{2} ln x^{2} + 3 + ln ∣ x ∣

= \frac{1}{2} ln x^{2} + 3 + ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 3 + ln ∣ x ∣ + C

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

\int \frac{9 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} d x

\int d (1 0^{x}) d x

\int \frac{x}{2 - x ^{4}} d x

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