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Popular >

integral of 1/(2+x)

Solution

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

Solution

ln ∣ 2 + x ∣ + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 2 + x

= ln ∣ 2 + x ∣

Add a constant to the solution

= ln ∣ 2 + x ∣ + C

Popular Examples

t^2y^{''}+ty^'+25y=0

t^{2} y^{^{''}} + t y^{^{'}} + 25 y = 0

\frac{d y}{d x} = 3 x

limit as x approaches-1 of-x^2+1

x \to - 1 lim (- x^{2} + 1)

y^{''}-12y^'+36y=0,y(0)=-2,y^'(0)=3

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 3

integral from 0 to 1 of 2/(t^2+1)

\int_{0}^{1} \frac{2}{t ^{2} + 1} d t

Frequently Asked Questions (FAQ)

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