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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u ^{\frac{3}{2}}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{2}{u ^{\frac{1}{2}}}

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

Substitute back

u = 1 - x^{2}

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

Simplify

- \frac{1}{2} (- \frac{2}{( 1 - x ^{2} ) ^{\frac{1}{2}}}) : \frac{1}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{x}{x ^{2} + y ^{2}})

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\frac{\partial}{\partial z} (3)

sin (x^{2})

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