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Popular >

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

Solution

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

Solution

2 ln ∣ y + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

= \int \frac{2}{u + 1} 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}{u + 1} 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 ∣ y + 1 ∣

Add a constant to the solution

= 2 ln ∣ y + 1 ∣ + C

Popular Examples

y^{^{'}} = sin (x^{2})

sum from n=0 to infinity of (n+4)/(3n+2)

n = 0 \sum \infty \frac{n + 4}{3 n + 2}

integral of (z-3)^n

\int (z - 3)^{n} d z

integral from-1 to 1 of (7x^3-2x^2+5x-4)

\int_{- 1}^{1} (7 x^{3} - 2 x^{2} + 5 x - 4) d x

(dy)/(dx)=(2x+1)/(2y)

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

Frequently Asked Questions (FAQ)

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