What is the integral of 1/((x^2+36)^{3/2)} ?

The integral of 1/((x^2+36)^{3/2)} is x/(36(36+x^2)^{1/2)}+C

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integral of 1/((x^2+36)^{3/2)}

\int \frac{1}{( x ^{2} + 36 ) ^{\frac{3}{2}}} d x

\frac{x}{36 ( 36 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} + 36 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{36 sec ( u )} d u

= \int \frac{1}{36 sec ( u )} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{36} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{36} sin (u)

Substitute back

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

= \frac{1}{36} sin (arctan (\frac{1}{6} x))

Simplify

\frac{1}{36} sin (arctan (\frac{1}{6} x)) : \frac{x}{36 ( 36 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{36 ( 36 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{36 ( 36 + x ^{2} ) ^{\frac{1}{2}}} + C

\frac{\partial}{\partial y} ((t^{2} + y^{2})^{\frac{3}{2}})

f (x) = (2 x + 4)^{\frac{1}{5}}, at x = 1

y \to \infty lim (\frac{( y ^{2} - e ^{(6 y)} )}{( 4 y ^{2} + e ^{7 y} )})

h (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 1)

What is the integral of 1/((x^2+36)^{3/2)} ?
The integral of 1/((x^2+36)^{3/2)} is x/(36(36+x^2)^{1/2)}+C