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

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

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

\int \frac{- 2 y}{( 1 - y ^{2} ) ^{\frac{1}{2}}} d y

- 2 (\frac{y ^{2}}{( 1 - y ^{2} ) ^{\frac{1}{2}}} - \frac{1}{( 1 - y ^{2} ) ^{\frac{1}{2}}}) + C

Solution steps

\int \frac{- 2 y}{( 1 - y ^{2} ) ^{\frac{1}{2}}} d y

Take the constant out:

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

= - 2 \cdot \int \frac{y}{( 1 - y ^{2} ) ^{\frac{1}{2}}} d y

Apply Integration By Parts

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

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

= - 2 (\frac{y ^{2}}{( 1 - y ^{2} ) ^{\frac{1}{2}}} - \frac{1}{( 1 - y ^{2} ) ^{\frac{1}{2}}})

Add a constant to the solution

= - 2 (\frac{y ^{2}}{( 1 - y ^{2} ) ^{\frac{1}{2}}} - \frac{1}{( 1 - y ^{2} ) ^{\frac{1}{2}}}) + C

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