What is the integral of (x^2+8x-3)/(x^3+3x) ?

The integral of (x^2+8x-3)/(x^3+3x) is ln|x^2+3|+8/(sqrt(3))arctan(x/(sqrt(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 (x^2+8x-3)/(x^3+3x)

\int \frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x} d x

ln x^{2} + 3 + \frac{8}{3} arctan (\frac{x}{3}) - ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x} d x

Expand

\frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x} : \frac{x ^{2}}{x ^{3} + 3 x} + \frac{8 x}{x ^{3} + 3 x} - \frac{3}{x ^{3} + 3 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 ^{2}}{x ^{3} + 3 x} d x + \int \frac{8 x}{x ^{3} + 3 x} d x - \int \frac{3}{x ^{3} + 3 x} d x

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

\int \frac{8 x}{x ^{3} + 3 x} d x = \frac{8}{3} arctan (\frac{x}{3})

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

= \frac{1}{2} ln x^{2} + 3 + \frac{8}{3} arctan (\frac{x}{3}) - (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 3)

Simplify

\frac{1}{2} ln x^{2} + 3 + \frac{8}{3} arctan (\frac{x}{3}) - (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 3) : ln x^{2} + 3 + \frac{8}{3} arctan (\frac{x}{3}) - ln ∣ x ∣

= ln x^{2} + 3 + \frac{8}{3} arctan (\frac{x}{3}) - ln ∣ x ∣

Add a constant to the solution

= ln x^{2} + 3 + \frac{8}{3} arctan (\frac{x}{3}) - ln ∣ x ∣ + C

y^{^{''}} + 5 y^{^{'}} + 6 y = 108 x^{2}

\int_{7}^{\infty} \frac{ln ( x )}{x} d x

\int_{1}^{2} (x^{2}) d x

\int 9 x^{- 3} d x

What is the integral of (x^2+8x-3)/(x^3+3x) ?
The integral of (x^2+8x-3)/(x^3+3x) is ln|x^2+3|+8/(sqrt(3))arctan(x/(sqrt(3)))-ln|x|+C