What is the integral of x/((9+4x^2)^{3/2)} ?

The integral of x/((9+4x^2)^{3/2)} is -1/(4(4x^2+9)^{1/2)}+C

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integral of x/((9+4x^2)^{3/2)}

\int \frac{x}{( 9 + 4 x ^{2} ) ^{\frac{3}{2}}} d x

- \frac{1}{4 ( 4 x ^{2} + 9 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x}{( 9 + 4 x ^{2} ) ^{\frac{3}{2}}} d x

Apply u-substitution

= \int \frac{1}{8 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}{8} \cdot \int \frac{1}{u ^{\frac{3}{2}}} d u

Apply the Power Rule

= \frac{1}{8} (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = 9 + 4 x^{2}

= \frac{1}{8} (- \frac{2}{( 9 + 4 x ^{2} ) ^{\frac{1}{2}}})

Simplify

\frac{1}{8} (- \frac{2}{( 9 + 4 x ^{2} ) ^{\frac{1}{2}}}) : - \frac{1}{4 ( 4 x ^{2} + 9 ) ^{\frac{1}{2}}}

= - \frac{1}{4 ( 4 x ^{2} + 9 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{1}{4 ( 4 x ^{2} + 9 ) ^{\frac{1}{2}}} + C

What is the integral of x/((9+4x^2)^{3/2)} ?
The integral of x/((9+4x^2)^{3/2)} is -1/(4(4x^2+9)^{1/2)}+C