What is the integral of x/((16+x^2)^{3/2)} ?

The integral of x/((16+x^2)^{3/2)} is -1/((16+x^2)^{1/2)}+C

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integral of x/((16+x^2)^{3/2)}

\int \frac{x}{( 16 + x ^{2} ) ^{\frac{3}{2}}} d x

- \frac{1}{( 16 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x}{( 16 + 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 = 16 + x^{2}

= \frac{1}{2} (- \frac{2}{( 16 + x ^{2} ) ^{\frac{1}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{( 16 + x ^{2} ) ^{\frac{1}{2}}}) : - \frac{1}{( 16 + x ^{2} ) ^{\frac{1}{2}}}

= - \frac{1}{( 16 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{1}{( 16 + x ^{2} ) ^{\frac{1}{2}}} + C

y = ln (\frac{1 + sin ( x )}{1 - sin ( x )})

\frac{d}{d x} (x (3 x + 1)^{5})

x \to 1 lim (\frac{sin ( x - 1 )}{x})

\int \frac{ln ( 2 x )}{3 x} d x

f (x) = 3 sin (x), at x = - \frac{π}{4}

What is the integral of x/((16+x^2)^{3/2)} ?
The integral of x/((16+x^2)^{3/2)} is -1/((16+x^2)^{1/2)}+C