What is the integral of ((2x+3))/((3x+4)) ?

The integral of ((2x+3))/((3x+4)) is 2/3 x+1/9 ln|3x+4|+C

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integral of ((2x+3))/((3x+4))

integral

\frac{( 2 x + 3 )}{( 3 x + 4 )}

\frac{2}{3} x + \frac{1}{9} ln ∣ 3 x + 4 ∣ + C

Solution steps

\int \frac{2 x + 3}{3 x + 4} d x

Expand

\frac{2 x + 3}{3 x + 4} : \frac{2 x}{3 x + 4} + \frac{3}{3 x + 4}

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}{3 x + 4} d x + \int \frac{3}{3 x + 4} d x

\int \frac{2 x}{3 x + 4} d x = 2 (\frac{1}{3} x - \frac{4}{9} ln ∣ 3 x + 4 ∣)

\int \frac{3}{3 x + 4} d x = ln ∣ 3 x + 4 ∣

= 2 (\frac{1}{3} x - \frac{4}{9} ln ∣ 3 x + 4 ∣) + ln ∣ 3 x + 4 ∣

Simplify

2 (\frac{1}{3} x - \frac{4}{9} ln ∣ 3 x + 4 ∣) + ln ∣ 3 x + 4 ∣ : \frac{2}{3} x + \frac{1}{9} ln ∣ 3 x + 4 ∣

= \frac{2}{3} x + \frac{1}{9} ln ∣ 3 x + 4 ∣

Add a constant to the solution

= \frac{2}{3} x + \frac{1}{9} ln ∣ 3 x + 4 ∣ + C

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

2 x cos (x^{2})

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

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

(\frac{1}{x}) (cos (x) - sin (x))

What is the integral of ((2x+3))/((3x+4)) ?
The integral of ((2x+3))/((3x+4)) is 2/3 x+1/9 ln|3x+4|+C