What is the integral of (5x^2+7x+2)/((x^2+2)(x+4)) ?

The integral of (5x^2+7x+2)/((x^2+2)(x+4)) is ln|x^2+2|-1/(sqrt(2))arctan(x/(sqrt(2)))+3ln|x+4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (5x^2+7x+2)/((x^2+2)(x+4))

\int \frac{5 x ^{2} + 7 x + 2}{( x ^{2} + 2 ) ( x + 4 )} d x

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

Solution steps

\int \frac{5 x ^{2} + 7 x + 2}{( x ^{2} + 2 ) ( x + 4 )} d x

Take the partial fraction of

\frac{5 x ^{2} + 7 x + 2}{( x ^{2} + 2 ) ( x + 4 )} : \frac{2 x - 1}{x ^{2} + 2} + \frac{3}{x + 4}

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

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

\int \frac{3}{x + 4} d x = 3 ln ∣ x + 4 ∣

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (2 y x^{2} - cos (y))

\frac{d y}{d x} = sin (6 x)

\int 7 tan^{2} (x) d x

l a pl a ce t r an s f or m 6 e^{5 t} cos (2 t) - e^{7 t}

t an g e n t f (x) = x^{2} + 3, (- 5, 28)

What is the integral of (5x^2+7x+2)/((x^2+2)(x+4)) ?
The integral of (5x^2+7x+2)/((x^2+2)(x+4)) is ln|x^2+2|-1/(sqrt(2))arctan(x/(sqrt(2)))+3ln|x+4|+C