What is the integral of 1/(4(x+1)) ?

The integral of 1/(4(x+1)) is 1/4 ln|x+1|+C

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integral of 1/(4(x+1))

\int \frac{1}{4 ( x + 1 )} d x

\frac{1}{4} ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{1}{4 ( x + 1 )} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{x + 1} d x

Apply u-substitution

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x + 1

= \frac{1}{4} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ x + 1 ∣ + C

\int_{0}^{π} x cos (x^{2}) d x

y^{^{'}} = (2 y + 22)^{0.5}

f (x) = (x^{3} + 1) \cdot 3 x^{2}

t sin (9 t)

y = x^{2} - x, y = 0

What is the integral of 1/(4(x+1)) ?
The integral of 1/(4(x+1)) is 1/4 ln|x+1|+C