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

The integral of ((6x+1))/((2x-3)) is 3x-9/2+5ln|2x-3|+C

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

integral

\frac{( 6 x + 1 )}{( 2 x - 3 )}

3 x - \frac{9}{2} + 5 ln ∣ 2 x - 3 ∣ + C

Solution steps

\int \frac{6 x + 1}{2 x - 3} d x

Expand

\frac{6 x + 1}{2 x - 3} : \frac{6 x}{2 x - 3} + \frac{1}{2 x - 3}

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{6 x}{2 x - 3} d x + \int \frac{1}{2 x - 3} d x

\int \frac{6 x}{2 x - 3} d x = 3 x - \frac{9}{2} + \frac{9}{2} ln ∣ 2 x - 3 ∣

\int \frac{1}{2 x - 3} d x = \frac{1}{2} ln ∣ 2 x - 3 ∣

= 3 x - \frac{9}{2} + \frac{9}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln ∣ 2 x - 3 ∣

Add similar elements:

\frac{9}{2} ln ∣ 2 x - 3 ∣ + \frac{1}{2} ln ∣ 2 x - 3 ∣ = 5 ln ∣ 2 x - 3 ∣

= 3 x - \frac{9}{2} + 5 ln ∣ 2 x - 3 ∣

Add a constant to the solution

= 3 x - \frac{9}{2} + 5 ln ∣ 2 x - 3 ∣ + C

\frac{9}{( 3 x - 4 ) ^{2}}

x^{5} cos (3 x)

y^{4} (1 - y)^{3}

\frac{1}{( x ) ^{2}}

e^{1} + 2 x

What is the integral of ((6x+1))/((2x-3)) ?
The integral of ((6x+1))/((2x-3)) is 3x-9/2+5ln|2x-3|+C