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

The integral of 1/(x(x^2+3)) is 1/3 ln|x|-1/6 ln|x^2+3|+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(x^2+3))

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

\frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{3 x} d x = \frac{1}{3} ln ∣ x ∣

\int \frac{x}{3 ( x ^{2} + 3 )} d x = \frac{1}{6} ln x^{2} + 3

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3

Add a constant to the solution

= \frac{1}{3} ln ∣ x ∣ - \frac{1}{6} ln x^{2} + 3 + C

\int \frac{x ^{3} + 4 x ^{2} + 8 x + 5}{x ^{2} + 2 x + 1} d x

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

\int \frac{tan ^{5} ( ln ( x ) )}{x} d x

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

\int (\frac{1}{x - 1})^{7} d x

What is the integral of 1/(x(x^2+3)) ?
The integral of 1/(x(x^2+3)) is 1/3 ln|x|-1/6 ln|x^2+3|+C