What is the integral of 1/((sqrt(9x^2+4))^3) ?

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

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integral of 1/((sqrt(9x^2+4))^3)

\int \frac{1}{( 9 x ^{2} + 4 ) ^{3}} d x

\frac{x}{4 9 x ^{2} + 4} + C

Solution steps

\int \frac{1}{( 9 x ^{2} + 4 ) ^{3}} d x

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

= \int \frac{1}{( 9 x ^{2} + 4 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{12 sec ( u )} d u

= \int \frac{1}{12 sec ( u )} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{12} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{12} sin (u)

Substitute back

u = arctan (\frac{3}{2} x)

= \frac{1}{12} sin (arctan (\frac{3}{2} x))

Simplify

\frac{1}{12} sin (arctan (\frac{3}{2} x)) : \frac{x}{4 9 x ^{2} + 4}

= \frac{x}{4 9 x ^{2} + 4}

Add a constant to the solution

= \frac{x}{4 9 x ^{2} + 4} + C

\int 4 e^{\frac{x}{2}} x d x

\int x^{- \frac{4}{5}} d x

\int 7 sin (7 s) d s

\int x^{4} ln (7 x) d x

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

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