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Popular >

integral of 1/(3+y)

Solution

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

Solution

ln ∣ 3 + y ∣ + C

Solution steps

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

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 = 3 + y

= ln ∣ 3 + y ∣

Add a constant to the solution

= ln ∣ 3 + y ∣ + C

Popular Examples

integral from 0 to 1 of 1/(1+9x^2)

\int_{0}^{1} \frac{1}{1 + 9 x ^{2}} d x

(\partial)/(\partial x)(2x-4)

\frac{\partial}{\partial x} (2 x - 4)

(\partial)/(\partial y)(-sin(xy))

\frac{\partial}{\partial y} (- sin (x y))

(dP)/(dt)=0.2P-0.001P^2

\frac{d P}{d t} = 0.2 P - 0.001 P^{2}

derivative of-(2e^x/(x^3)+(e^x)/(x^2))

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

Frequently Asked Questions (FAQ)

What is the integral of 1/(3+y) ?
The integral of 1/(3+y) is ln|3+y|+C