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

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

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

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

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

Solution steps

\int \frac{x}{( x ^{2} + 4 ) ^{\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 = x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2} + 2, [- 2, 1]

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