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

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

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integral of 1/((9x^2+1)^{3/2)}

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Use the common integral:

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

= \frac{1}{3} sin (u)

Substitute back

u = arctan (3 x)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{u}{9} d u

\int - 5 x^{4} + \frac{- 8}{x} d x

\int 5^{\frac{1}{2}} d x

\int 2 x^{2} \cdot e^{x} d x

\int x (1 + x^{2}) arccot (x) d x

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