What is the integral of (12x)/(2x-5) ?

The integral of (12x)/(2x-5) is 3(2x-5+5ln|2x-5|)+C

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integral of (12x)/(2x-5)

\int \frac{12 x}{2 x - 5} d x

3 (2 x - 5 + 5 ln ∣ 2 x - 5 ∣) + C

Solution steps

\int \frac{12 x}{2 x - 5} d x

Take the constant out:

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

= 12 \cdot \int \frac{x}{2 x - 5} d x

Apply u-substitution

: \int \frac{u + 5}{4 u} d u

= 12 \cdot \int \frac{u + 5}{4 u} d u

Take the constant out:

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

= 12 \cdot \frac{1}{4} \cdot \int \frac{u + 5}{u} d u

Expand

\frac{u + 5}{u} : 1 + \frac{5}{u}

= 12 \cdot \frac{1}{4} \cdot \int 1 + \frac{5}{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

= 12 \cdot \frac{1}{4} (\int 1 d u + \int \frac{5}{u} d u)

\int 1 d u = u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= 12 \cdot \frac{1}{4} (u + 5 ln ∣ u ∣)

Substitute back

u = 2 x - 5

= 12 \cdot \frac{1}{4} (2 x - 5 + 5 ln ∣ 2 x - 5 ∣)

Simplify

12 \cdot \frac{1}{4} (2 x - 5 + 5 ln ∣ 2 x - 5 ∣) : 3 (2 x - 5 + 5 ln ∣ 2 x - 5 ∣)

= 3 (2 x - 5 + 5 ln ∣ 2 x - 5 ∣)

Add a constant to the solution

= 3 (2 x - 5 + 5 ln ∣ 2 x - 5 ∣) + C

x \to 0 lim (\frac{csc ( 2 x )}{x})

\frac{d}{d x} ((3 x - 5)^{7} (5 x + 2)^{4})

\frac{\partial}{\partial x} (t^{2} e^{\frac{x ^{2}}{4 t}})

\frac{\partial}{\partial x} (e^{- 2 (x + y)})

f (x) = - 64 x (- 16 x^{2} + 36)

What is the integral of (12x)/(2x-5) ?
The integral of (12x)/(2x-5) is 3(2x-5+5ln|2x-5|)+C