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

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

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

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

2 arctan (y) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= 2 arctan (u)

Substitute back

u = y

= 2 arctan (y)

Add a constant to the solution

= 2 arctan (y) + C

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