What is the integral of 6/(2x-5) ?

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

Popular >

integral of 6/(2x-5)

\int \frac{6}{2 x - 5} d x

3 ln ∣ 2 x - 5 ∣ + C

Solution steps

\int \frac{6}{2 x - 5} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{2 x - 5} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 6 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x - 5

= 6 \cdot \frac{1}{2} ln ∣ 2 x - 5 ∣

Simplify

6 \cdot \frac{1}{2} ln ∣ 2 x - 5 ∣ : 3 ln ∣ 2 x - 5 ∣

= 3 ln ∣ 2 x - 5 ∣

Add a constant to the solution

= 3 ln ∣ 2 x - 5 ∣ + C

n = 0 \sum \infty \frac{0 ^{n}}{3 ^{n}}

\frac{\partial}{\partial z} (z arctan (\frac{y}{x}))

\frac{d}{d s} (s - t)

\frac{( x + 1 ) ^{2}}{( 3 x - 4 )}

\int \frac{4 x ^{3} - 3 x}{( x ^{2} + 1 ) ^{2}} d x