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

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

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integral of 1/((3x+1))

\int \frac{1}{( 3 x + 1 )} d x

\frac{1}{3} ln ∣ 3 x + 1 ∣ + C

Solution steps

\int \frac{1}{3 x + 1} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 1

= \frac{1}{3} ln ∣ 3 x + 1 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x + 1 ∣ + C

x \to 3 + lim (\frac{x + 4}{x - 3})

f (x) = cot^{2} (cos (θ))

\int \frac{( 2 x - 5 ) ^{2}}{3} d x

\int_{0}^{\frac{π}{6}} 3 cos^{5} (3 x) d x

\frac{d}{d y} (x^{2} + x y + y^{2})

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