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

The integral of (5x+6)/((x-2)(3x+1)^2) is 16/49 ln|x-2|-16/49 ln|3x+1|+(13)/(21(3x+1))+C

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integral of (5x+6)/((x-2)(3x+1)^2)

\int \frac{5 x + 6}{( x - 2 ) ( 3 x + 1 ) ^{2}} d x

\frac{16}{49} ln ∣ x - 2 ∣ - \frac{16}{49} ln ∣ 3 x + 1 ∣ + \frac{13}{21 ( 3 x + 1 )} + C

Solution steps

\int \frac{5 x + 6}{( x - 2 ) ( 3 x + 1 ) ^{2}} d x

Take the partial fraction of

\frac{5 x + 6}{( x - 2 ) ( 3 x + 1 ) ^{2}} : \frac{16}{49 ( x - 2 )} - \frac{48}{49 ( 3 x + 1 )} - \frac{13}{7 ( 3 x + 1 ) ^{2}}

= \int \frac{16}{49 ( x - 2 )} - \frac{48}{49 ( 3 x + 1 )} - \frac{13}{7 ( 3 x + 1 ) ^{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{16}{49 ( x - 2 )} d x - \int \frac{48}{49 ( 3 x + 1 )} d x - \int \frac{13}{7 ( 3 x + 1 ) ^{2}} d x

\int \frac{16}{49 ( x - 2 )} d x = \frac{16}{49} ln ∣ x - 2 ∣

\int \frac{48}{49 ( 3 x + 1 )} d x = \frac{16}{49} ln ∣ 3 x + 1 ∣

\int \frac{13}{7 ( 3 x + 1 ) ^{2}} d x = - \frac{13}{21 ( 3 x + 1 )}

= \frac{16}{49} ln ∣ x - 2 ∣ - \frac{16}{49} ln ∣ 3 x + 1 ∣ - (- \frac{13}{21 ( 3 x + 1 )})

Simplify

= \frac{16}{49} ln ∣ x - 2 ∣ - \frac{16}{49} ln ∣ 3 x + 1 ∣ + \frac{13}{21 ( 3 x + 1 )}

Add a constant to the solution

= \frac{16}{49} ln ∣ x - 2 ∣ - \frac{16}{49} ln ∣ 3 x + 1 ∣ + \frac{13}{21 ( 3 x + 1 )} + C

f (y) = e^{t a n (θ)}

\frac{d y}{d x} = \frac{y + 4}{2 - y}

\frac{d q}{d t} = k (q - 70)

\frac{d}{d x} (6^{- x})

x^{3}

What is the integral of (5x+6)/((x-2)(3x+1)^2) ?
The integral of (5x+6)/((x-2)(3x+1)^2) is 16/49 ln|x-2|-16/49 ln|3x+1|+(13)/(21(3x+1))+C