What is the integral of 1/(4x-10) ?

The integral of 1/(4x-10) is 1/4 ln|4x-10|+C

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integral of 1/(4x-10)

\int \frac{1}{4 x - 10} d x

\frac{1}{4} ln ∣ 4 x - 10 ∣ + C

Solution steps

\int \frac{1}{4 x - 10} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 4 x - 10

= \frac{1}{4} ln ∣ 4 x - 10 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x - 10 ∣ + C

\frac{\partial}{\partial x} (sin^{2} (θ))

x = - 5, x = 1, y = 2 x^{2} + 4, y = 0

\int_{- 1}^{1} x + 1 d x

x \to 0 + lim (\frac{1}{x})

\int_{0}^{\frac{3 π}{2}} cos^{3} (33 x) d x

What is the integral of 1/(4x-10) ?
The integral of 1/(4x-10) is 1/4 ln|4x-10|+C