What is the integral of (20x^2+x+45)/(x(4x^2+9)) ?

The integral of (20x^2+x+45)/(x(4x^2+9)) is 5ln|x|+1/6 arctan(2/3 x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (20x^2+x+45)/(x(4x^2+9))

\int \frac{20 x ^{2} + x + 45}{x ( 4 x ^{2} + 9 )} d x

5 ln ∣ x ∣ + \frac{1}{6} arctan (\frac{2}{3} x) + C

Solution steps

\int \frac{20 x ^{2} + x + 45}{x ( 4 x ^{2} + 9 )} d x

Take the partial fraction of

\frac{20 x ^{2} + x + 45}{x ( 4 x ^{2} + 9 )} : \frac{5}{x} + \frac{1}{4 x ^{2} + 9}

= \int \frac{5}{x} + \frac{1}{4 x ^{2} + 9} 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{5}{x} d x + \int \frac{1}{4 x ^{2} + 9} d x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

\int \frac{1}{4 x ^{2} + 9} d x = \frac{1}{6} arctan (\frac{2}{3} x)

= 5 ln ∣ x ∣ + \frac{1}{6} arctan (\frac{2}{3} x)

Add a constant to the solution

= 5 ln ∣ x ∣ + \frac{1}{6} arctan (\frac{2}{3} x) + C

y = \frac{4}{( 2 x ) ^{3}}

\int (- 7 - 7 tan^{2} (θ)) d θ

\frac{d P}{d x} = 0.05 P (1 - \frac{P}{2000})

\frac{\partial}{\partial x} (5 x + 8 y + 10 z)

f^{^{'}} (x) = \frac{x ^{2}}{( 9 + 4 x )}

What is the integral of (20x^2+x+45)/(x(4x^2+9)) ?
The integral of (20x^2+x+45)/(x(4x^2+9)) is 5ln|x|+1/6 arctan(2/3 x)+C