What is the integral of ((16x^2))/((sqrt(x^2+9))^5) ?

The integral of ((16x^2))/((sqrt(x^2+9))^5) is (16x^3)/(27(9+x^2)^{3/2)}+C

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integral of ((16x^2))/((sqrt(x^2+9))^5)

\int \frac{( 16 x ^{2} )}{( x ^{2} + 9 ) ^{5}} d x

\frac{16 x ^{3}}{27 ( 9 + x ^{2} ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{16 x ^{2}}{( x ^{2} + 9 ) ^{5}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{2}}{( x ^{2} + 9 ) ^{5}} d x

(x^{2} + 9)^{5} = (x^{2} + 9)^{\frac{5}{2}}

= 16 \cdot \int \frac{x ^{2}}{( x ^{2} + 9 ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

= 16 \cdot \int \frac{tan ^{2} ( u )}{9 sec ^{3} ( u )} d u

Take the constant out:

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

= 16 \cdot \frac{1}{9} \cdot \int \frac{tan ^{2} ( u )}{sec ^{3} ( u )} d u

Rewrite using trig identities

= 16 \cdot \frac{1}{9} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

= 16 \cdot \frac{1}{9} \cdot \int v^{2} d v

Apply the Power Rule

= 16 \cdot \frac{1}{9} \cdot \frac{v ^{3}}{3}

Substitute back

= 16 \cdot \frac{1}{9} \cdot \frac{sin ^{3} ( arctan ( \frac{1}{3} x ) )}{3}

Simplify

16 \cdot \frac{1}{9} \cdot \frac{sin ^{3} ( arctan ( \frac{1}{3} x ) )}{3} : \frac{16 x ^{3}}{27 ( 9 + x ^{2} ) ^{\frac{3}{2}}}

= \frac{16 x ^{3}}{27 ( 9 + x ^{2} ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{16 x ^{3}}{27 ( 9 + x ^{2} ) ^{\frac{3}{2}}} + C

What is the integral of ((16x^2))/((sqrt(x^2+9))^5) ?
The integral of ((16x^2))/((sqrt(x^2+9))^5) is (16x^3)/(27(9+x^2)^{3/2)}+C