What is the integral of x/((4-x^2)^{3/2)} ?

The integral of x/((4-x^2)^{3/2)} is 1/((4-x^2)^{1/2)}+C

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integral of x/((4-x^2)^{3/2)}

\int \frac{x}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{1}{( 4 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x}{( 4 - 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 = 4 - x^{2}

= - \frac{1}{2} (- \frac{2}{( 4 - x ^{2} ) ^{\frac{1}{2}}})

Simplify

- \frac{1}{2} (- \frac{2}{( 4 - x ^{2} ) ^{\frac{1}{2}}}) : \frac{1}{( 4 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{1}{( 4 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{1}{( 4 - x ^{2} ) ^{\frac{1}{2}}} + C

\int 5 x^{\frac{1}{4}} - 7 x^{\frac{3}{4}} d x

f (x) = \frac{7}{x}

\int \frac{3 x ^{4}}{x ^{2} - 2 x} d x

\frac{- x ^{2} + x + 6}{e ^{x}}

ln (x^{2} - x + 1)

What is the integral of x/((4-x^2)^{3/2)} ?
The integral of x/((4-x^2)^{3/2)} is 1/((4-x^2)^{1/2)}+C