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

The integral of x/(2x^2+5x+2) is -1/6 ln|2x+1|+2/3 ln|x+2|+C

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

\int \frac{x}{2 x ^{2} + 5 x + 2} d x

- \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{x}{2 x ^{2} + 5 x + 2} d x

Take the partial fraction of

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

= \int - \frac{1}{3 ( 2 x + 1 )} + \frac{2}{3 ( x + 2 )} 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{1}{3 ( 2 x + 1 )} d x + \int \frac{2}{3 ( x + 2 )} d x

\int \frac{1}{3 ( 2 x + 1 )} d x = \frac{1}{6} ln ∣ 2 x + 1 ∣

\int \frac{2}{3 ( x + 2 )} d x = \frac{2}{3} ln ∣ x + 2 ∣

= - \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{1}{6} ln ∣ 2 x + 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣ + C

x \to \infty lim (- 28 x^{4})

\frac{d y}{d x}, x y^{2} = 4

\int \frac{2 x}{( x + 2 ) ( x + 1 )} d x

\frac{\partial}{\partial x} (4 x - 2 y)

y = 4^{x}, (1, 4)

What is the integral of x/(2x^2+5x+2) ?
The integral of x/(2x^2+5x+2) is -1/6 ln|2x+1|+2/3 ln|x+2|+C