What is the integral of 1/((x^2+16)^{3/2)} ?

The integral of 1/((x^2+16)^{3/2)} is x/(16(16+x^2)^{1/2)}+C

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integral of 1/((x^2+16)^{3/2)}

\int \frac{1}{( x ^{2} + 16 ) ^{\frac{3}{2}}} d x

\frac{x}{16 ( 16 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( x ^{2} + 16 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 sec ( u )} d u

= \int \frac{1}{16 sec ( u )} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{16} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{16} sin (u)

Substitute back

u = arctan (\frac{1}{4} x)

= \frac{1}{16} sin (arctan (\frac{1}{4} x))

Simplify

\frac{1}{16} sin (arctan (\frac{1}{4} x)) : \frac{x}{16 ( 16 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{16 ( 16 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{16 ( 16 + x ^{2} ) ^{\frac{1}{2}}} + C

f (x) = x^{2} - 3, at x = 3

\int_{a}^{2 a} (a x - \frac{a ^{3}}{x}) d x

x \to \infty lim (4 e^{- 4 x})

\int x^{\frac{6}{5}} d x

\int_{1}^{3} 3 (2^{x + 1}) d x

What is the integral of 1/((x^2+16)^{3/2)} ?
The integral of 1/((x^2+16)^{3/2)} is x/(16(16+x^2)^{1/2)}+C