What is the integral of 1/((324+x^2)^{3/2)} ?

The integral of 1/((324+x^2)^{3/2)} is x/(324(324+x^2)^{1/2)}+C

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integral of 1/((324+x^2)^{3/2)}

\int \frac{1}{( 324 + x ^{2} ) ^{\frac{3}{2}}} d x

\frac{x}{324 ( 324 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 324 + x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{324 sec ( u )} d u

= \int \frac{1}{324 sec ( u )} d u

Take the constant out:

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

= \frac{1}{324} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{324} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{324} sin (u)

Substitute back

u = arctan (\frac{1}{18} x)

= \frac{1}{324} sin (arctan (\frac{1}{18} x))

Simplify

\frac{1}{324} sin (arctan (\frac{1}{18} x)) : \frac{x}{324 ( 324 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{324 ( 324 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{324 ( 324 + x ^{2} ) ^{\frac{1}{2}}} + C

2 tan^{2} (2 x)

\frac{d}{d t} (ln (t^{2} + 1))

y^{^{'}} = y (1 - y), y (0) = 6

y^{^{'}} = 1 - y^{2} sin (x)

x \to 0 + lim (\frac{x e ^{\frac{1}{x}}}{2})

What is the integral of 1/((324+x^2)^{3/2)} ?
The integral of 1/((324+x^2)^{3/2)} is x/(324(324+x^2)^{1/2)}+C