What is the integral of 1/((5x+1)) ?

The integral of 1/((5x+1)) is 1/5 ln|5x+1|+C

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integral of 1/((5x+1))

\int \frac{1}{( 5 x + 1 )} d x

\frac{1}{5} ln ∣ 5 x + 1 ∣ + C

Solution steps

\int \frac{1}{5 x + 1} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= \int \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} ln ∣ u ∣

Substitute back

u = 5 x + 1

= \frac{1}{5} ln ∣ 5 x + 1 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ 5 x + 1 ∣ + C

(t + 1)^{2} e^{- 2 t}

(y^{2} - x y) d x + (x^{2}) d y = 0

(3.1) (4.2)

\frac{d}{d x} (\frac{e ^{x}}{6 x ^{2} + 5})

\frac{\partial}{\partial x} (x^{2} + y^{2} + x^{2} y + 9)

What is the integral of 1/((5x+1)) ?
The integral of 1/((5x+1)) is 1/5 ln|5x+1|+C