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

The integral of (2x+5)/(3x-1) is 1/9 (2(3x-1)+17ln|3x-1|)+C

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

\int \frac{2 x + 5}{3 x - 1} d x

\frac{1}{9} (2 (3 x - 1) + 17 ln ∣ 3 x - 1 ∣) + C

Solution steps

\int \frac{2 x + 5}{3 x - 1} d x

Apply u-substitution

: \int \frac{2 u + 17}{9 u} d u

= \int \frac{2 u + 17}{9 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int \frac{2 u + 17}{u} d u

Expand

\frac{2 u + 17}{u} : 2 + \frac{17}{u}

= \frac{1}{9} \cdot \int 2 + \frac{17}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{9} (\int 2 d u + \int \frac{17}{u} d u)

\int 2 d u = 2 u

\int \frac{17}{u} d u = 17 ln ∣ u ∣

= \frac{1}{9} (2 u + 17 ln ∣ u ∣)

Substitute back

u = 3 x - 1

= \frac{1}{9} (2 (3 x - 1) + 17 ln ∣ 3 x - 1 ∣)

Add a constant to the solution

= \frac{1}{9} (2 (3 x - 1) + 17 ln ∣ 3 x - 1 ∣) + C

\frac{d}{d x} ((\frac{1}{3 x + 1})^{4})

\frac{4}{s ^{2} + 9}

\frac{d y}{d x} = 1 - 2 y, y (0) = 0

\int 11 sin^{5} (x) d x

\frac{d y}{d x} = (\frac{( 2 y + 7 )}{( 4 x + 9 )})^{2}

What is the integral of (2x+5)/(3x-1) ?
The integral of (2x+5)/(3x-1) is 1/9 (2(3x-1)+17ln|3x-1|)+C