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

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

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integral of 1/(1+3x)

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

\frac{1}{3} ln ∣ 1 + 3 x ∣ + C

Solution steps

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

= \frac{1}{3} ln ∣ 1 + 3 x ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 1 + 3 x ∣ + C

x \to \infty lim (x^{2} + 4 x)

(x (ln (x))^{2})^{^{'}}

\int \frac{1}{( ρ ^{2} + z ^{2} ) ^{\frac{3}{2}}} d z

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

sin (y) \frac{d y}{d x} - cos (y) = 0