What is the integral of (y^{1/2})/(1-y^{3/2)} ?

The integral of (y^{1/2})/(1-y^{3/2)} is -2/3 ln|1-y^{3/2}|+C

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integral of (y^{1/2})/(1-y^{3/2)}

\int \frac{y ^{\frac{1}{2}}}{1 - y ^{\frac{3}{2}}} d y

- \frac{2}{3} ln 1 - y^{\frac{3}{2}} + C

Solution steps

\int \frac{y ^{\frac{1}{2}}}{1 - y ^{\frac{3}{2}}} d y

Apply u-substitution

: \int - \frac{2}{3 u} d u

= \int - \frac{2}{3 u} d u

Take the constant out:

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

= - \frac{2}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{2}{3} ln ∣ u ∣

Substitute back

u = 1 - y^{\frac{3}{2}}

= - \frac{2}{3} ln 1 - y^{\frac{3}{2}}

Add a constant to the solution

= - \frac{2}{3} ln 1 - y^{\frac{3}{2}} + C

\frac{12 s + 5}{( s + 3 ) ^{2} ( s + 7 )}

\int 2 + 2 x d x

\int_{2}^{3} \frac{- 5}{x ^{2} - 1} d x

\frac{d}{d x} (\frac{x + 6}{7 x ^{2} e ^{x}})

\int e^{- \frac{x}{5}} d x

What is the integral of (y^{1/2})/(1-y^{3/2)} ?
The integral of (y^{1/2})/(1-y^{3/2)} is -2/3 ln|1-y^{3/2}|+C