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

The integral of (3x+1)/(x^3+4x^2+5x) is 1/5 ln|x|+13/5 arctan(x+2)-1/10 ln|x^2+4x+5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x+1)/(x^3+4x^2+5x)

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

\frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5 + C

Solution steps

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

Expand

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

\int \frac{3 x}{x ^{3} + 4 x ^{2} + 5 x} d x = 3 arctan (x + 2)

\int \frac{1}{x ^{3} + 4 x ^{2} + 5 x} d x = \frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} + 4 x + 5 - 4 arctan (x + 2))

= 3 arctan (x + 2) + \frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} + 4 x + 5 - 4 arctan (x + 2))

Simplify

3 arctan (x + 2) + \frac{1}{5} ln ∣ x ∣ + \frac{1}{10} (- ln x^{2} + 4 x + 5 - 4 arctan (x + 2)) : \frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5

= \frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5

Add a constant to the solution

= \frac{1}{5} ln ∣ x ∣ + \frac{13}{5} arctan (x + 2) - \frac{1}{10} ln x^{2} + 4 x + 5 + C

\int \frac{sin ( 4 x )}{sin ( 2 x ) cos ^{2} ( 2 x )} d x

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

\int 2 sec^{6} (10 x) d x

\int b 1 - \frac{x ^{2}}{a ^{2}} d x

\int 4 sin (5 x + 7) d x

What is the integral of (3x+1)/(x^3+4x^2+5x) ?
The integral of (3x+1)/(x^3+4x^2+5x) is 1/5 ln|x|+13/5 arctan(x+2)-1/10 ln|x^2+4x+5|+C