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

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

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integral of 2/(x+1)

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

2 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

u = x + 1

= 2 ln ∣ x + 1 ∣

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ + C

y^{^{'}} = \frac{x y ^{2} - 0.5 sin ( 2 x )}{y ( 1 - x ^{2} )}, y (0) = 2

y^{^{'}} = (t + y)^{2} - 1, y (2) = 6

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

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

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