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

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

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integral of 1/((x^2+6)^{3/2)}

\int \frac{1}{( x ^{2} + 6 ) ^{\frac{3}{2}}} d x

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

Solution steps

\int \frac{1}{( x ^{2} + 6 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{6 sec ( u )} d u

= \int \frac{1}{6 sec ( u )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{6} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{6} sin (u)

Substitute back

u = arctan (\frac{1}{6 ^{\frac{1}{2}}} x)

= \frac{1}{6} sin (arctan (\frac{1}{6 ^{\frac{1}{2}}} x))

Simplify

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

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

Add a constant to the solution

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

n = 0 \sum \infty 2 x^{n}

\frac{x}{x ^{2} + 36}

x \to 0 lim ((x)^{x})

y^{^{'}} = \frac{5 x}{y}

\frac{\partial}{\partial x} (e^{x \cdot y})

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