What is the integral of (4x+7)/(3x-1) ?

The integral of (4x+7)/(3x-1) is 1/9 (4(3x-1)+25ln|3x-1|)+C

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integral of (4x+7)/(3x-1)

\int \frac{4 x + 7}{3 x - 1} d x

\frac{1}{9} (4 (3 x - 1) + 25 ln ∣ 3 x - 1 ∣) + C

Solution steps

\int \frac{4 x + 7}{3 x - 1} d x

Apply u-substitution

: \int \frac{4 u + 25}{9 u} d u

= \int \frac{4 u + 25}{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{4 u + 25}{u} d u

Expand

\frac{4 u + 25}{u} : 4 + \frac{25}{u}

= \frac{1}{9} \cdot \int 4 + \frac{25}{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 4 d u + \int \frac{25}{u} d u)

\int 4 d u = 4 u

\int \frac{25}{u} d u = 25 ln ∣ u ∣

= \frac{1}{9} (4 u + 25 ln ∣ u ∣)

Substitute back

u = 3 x - 1

= \frac{1}{9} (4 (3 x - 1) + 25 ln ∣ 3 x - 1 ∣)

Add a constant to the solution

= \frac{1}{9} (4 (3 x - 1) + 25 ln ∣ 3 x - 1 ∣) + C

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What is the integral of (4x+7)/(3x-1) ?
The integral of (4x+7)/(3x-1) is 1/9 (4(3x-1)+25ln|3x-1|)+C