What is the integral of+1/((9x^2+1)^{5/2)} ?

The integral of+1/((9x^2+1)^{5/2)} is (x(6x^2+1))/((9x^2+1)^{3/2)}+C

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integral of+1/((9x^2+1)^{5/2)}

\int + \frac{1}{( 9 x ^{2} + 1 ) ^{\frac{5}{2}}} d x

\frac{x ( 6 x ^{2} + 1 )}{( 9 x ^{2} + 1 ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{1}{( 9 x ^{2} + 1 ) ^{\frac{5}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{3 sec ^{3} ( u )} d u

= \int \frac{1}{3 sec ^{3} ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{sec ^{3} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - sin^{2} (u)) cos (u) d u

Apply u-substitution

: \int 1 - v^{2} d v

= \frac{1}{3} \cdot \int 1 - v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 1 d v - \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{3} (v - \frac{v ^{3}}{3})

Substitute back

= \frac{1}{3} (sin (arctan (3 x)) - \frac{sin ^{3} ( arctan ( 3 x ) )}{3})

Simplify

\frac{1}{3} (sin (arctan (3 x)) - \frac{sin ^{3} ( arctan ( 3 x ) )}{3}) : \frac{x ( 6 x ^{2} + 1 )}{( 9 x ^{2} + 1 ) ^{\frac{3}{2}}}

= \frac{x ( 6 x ^{2} + 1 )}{( 9 x ^{2} + 1 ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{x ( 6 x ^{2} + 1 )}{( 9 x ^{2} + 1 ) ^{\frac{3}{2}}} + C

f (x) = \frac{6 x}{( x ^{3} - 9 ) ^{2}}

x \to 3 + lim (2 x + 5)

x \to 0 lim (\frac{x}{5 - x - 5 + x})

\frac{d}{d x} (ln (a + x 2 a x + x^{2}))

x \to 3 lim (f (x) x)

What is the integral of+1/((9x^2+1)^{5/2)} ?
The integral of+1/((9x^2+1)^{5/2)} is (x(6x^2+1))/((9x^2+1)^{3/2)}+C