What is the integral of 1/(10x-3) ?

The integral of 1/(10x-3) is 1/10 ln|10x-3|+C

Popular >

integral of 1/(10x-3)

\int \frac{1}{10 x - 3} d x

\frac{1}{10} ln ∣ 10 x - 3 ∣ + C

Solution steps

\int \frac{1}{10 x - 3} d x

Apply u-substitution

: \int \frac{1}{10 u} d u

= \int \frac{1}{10 u} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{10} ln ∣ u ∣

Substitute back

u = 10 x - 3

= \frac{1}{10} ln ∣ 10 x - 3 ∣

Add a constant to the solution

= \frac{1}{10} ln ∣ 10 x - 3 ∣ + C

y = (3 x - 1) (- 3 x^{2} - 4)^{- 3}

\int sec^{2} (x) sec (x) d x

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 1

- \frac{x ^{3}}{9}

\int \frac{x - 3}{x ^{2} - x - 6} d x

What is the integral of 1/(10x-3) ?
The integral of 1/(10x-3) is 1/10 ln|10x-3|+C