What is the integral of 1/((w^2+5)^{3/2)} ?

The integral of 1/((w^2+5)^{3/2)} is w/(5(5+w^2)^{1/2)}+C

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integral of 1/((w^2+5)^{3/2)}

\int \frac{1}{( w ^{2} + 5 ) ^{\frac{3}{2}}} d w

\frac{w}{5 ( 5 + w ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( w ^{2} + 5 ) ^{\frac{3}{2}}} d w

Apply Trigonometric Substitution

: \int \frac{1}{5 sec ( u )} d u

= \int \frac{1}{5 sec ( u )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{5} sin (u)

Substitute back

u = arctan (\frac{1}{5 ^{\frac{1}{2}}} w)

= \frac{1}{5} sin (arctan (\frac{1}{5 ^{\frac{1}{2}}} w))

Simplify

\frac{1}{5} sin (arctan (\frac{1}{5 ^{\frac{1}{2}}} w)) : \frac{w}{5 ( 5 + w ^{2} ) ^{\frac{1}{2}}}

= \frac{w}{5 ( 5 + w ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{w}{5 ( 5 + w ^{2} ) ^{\frac{1}{2}}} + C

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

n = 1 \sum \infty \frac{n + 1}{2 ^{n}}

\int_{0}^{π} 4 sin^{4} (x) d x

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

x^{2} - 8 x

What is the integral of 1/((w^2+5)^{3/2)} ?
The integral of 1/((w^2+5)^{3/2)} is w/(5(5+w^2)^{1/2)}+C