What is the integral of x/((x+2)(4x-1)) ?

The integral of x/((x+2)(4x-1)) is 2/9 ln|x+2|+1/36 ln|4x-1|+C

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integral of x/((x+2)(4x-1))

\int \frac{x}{( x + 2 ) ( 4 x - 1 )} d x

\frac{2}{9} ln ∣ x + 2 ∣ + \frac{1}{36} ln ∣ 4 x - 1 ∣ + C

Solution steps

\int \frac{x}{( x + 2 ) ( 4 x - 1 )} d x

Take the partial fraction of

\frac{x}{( x + 2 ) ( 4 x - 1 )} : \frac{2}{9 ( x + 2 )} + \frac{1}{9 ( 4 x - 1 )}

= \int \frac{2}{9 ( x + 2 )} + \frac{1}{9 ( 4 x - 1 )} 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}{9 ( x + 2 )} d x + \int \frac{1}{9 ( 4 x - 1 )} d x

\int \frac{2}{9 ( x + 2 )} d x = \frac{2}{9} ln ∣ x + 2 ∣

\int \frac{1}{9 ( 4 x - 1 )} d x = \frac{1}{36} ln ∣ 4 x - 1 ∣

= \frac{2}{9} ln ∣ x + 2 ∣ + \frac{1}{36} ln ∣ 4 x - 1 ∣

Add a constant to the solution

= \frac{2}{9} ln ∣ x + 2 ∣ + \frac{1}{36} ln ∣ 4 x - 1 ∣ + C

\frac{d}{d x} (x^{2.1})

\frac{d}{d x} (ln (\frac{a ^{2} - x ^{2}}{x}))

y (X) = - 3^{X}

x \to 3 lim (2 x^{2} + 1)

\frac{d y}{d x} + \frac{2}{x} y = x^{2} y^{3}

What is the integral of x/((x+2)(4x-1)) ?
The integral of x/((x+2)(4x-1)) is 2/9 ln|x+2|+1/36 ln|4x-1|+C