What is the integral of 1/(sqrt(y)+y) ?

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

Popular >

integral of 1/(sqrt(y)+y)

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

2 ln ∣ 1 + y ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{1 + u} d u

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 ln ∣ v ∣

Substitute back

= 2 ln ∣ 1 + y ∣

Add a constant to the solution

= 2 ln ∣ 1 + y ∣ + C

\int \frac{x ^{3}}{x ^{2} + 30 x + 225} d x

t \to 2 lim (\frac{2 t}{4 - t ^{2}})

\frac{d y}{d x} = - 3 + \frac{2}{1} \frac{d y}{d x}

f (x) = 15 x ln (4 x)

\frac{d}{d x} (\frac{7 - x e ^{x}}{x + e ^{x}})

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