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

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

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integral of 3/(1+x)

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

3 ln ∣ 1 + x ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 ln ∣ u ∣

Substitute back

u = 1 + x

= 3 ln ∣ 1 + x ∣

Add a constant to the solution

= 3 ln ∣ 1 + x ∣ + C

\frac{d}{d x} (e^{x^{3 + 6 x^{2}}})

\frac{d y}{d t} = \frac{y}{1 + t} - \frac{y}{t} + t^{2} (1 + t)

\frac{\partial}{\partial x} ((13 x - 1) (3 y - 1))

y = ln (3 x - 7)

\int \frac{2 x ^{3} - x ^{2} + 2 x + 2}{x ^{2} + 1} d x