What is the integral of 1/(2(x+1)) ?

The integral of 1/(2(x+1)) is 1/2 ln|x+1|+C

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integral of 1/(2(x+1))

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

\frac{1}{2} ln ∣ x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{x + 1} d x

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x + 1

= \frac{1}{2} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 1 ∣ + C

f (x) = x - x

\int \frac{( x + 7 )}{x ^{2} + 9} d x

n = 1 \sum \infty \frac{x ^{n}}{n}

\int \frac{1}{( 64 + x ^{2} ) ^{2}} d x

\frac{d y}{d x} + 2 y = x^{2} + 2 x

What is the integral of 1/(2(x+1)) ?
The integral of 1/(2(x+1)) is 1/2 ln|x+1|+C