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

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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

\int 2 e^{\frac{r}{2}} d r

\frac{\partial}{\partial x} (3 x x^{2} + y^{2})

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

\int \frac{1}{x ^{2} 225 - x ^{2}} d x

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

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