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

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

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

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

\frac{x}{16 16 + x ^{2}} + C

Solution steps

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

(16 + x^{2})^{3} = (16 + x^{2})^{\frac{3}{2}},

assuming

(16 + x^{2}) \geq 0

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

Add a constant to the solution

= \frac{x}{16 16 + x ^{2}} + C

- 6 x = 10 - y - 36 x = 48 - 6 y

\int 4 cos^{2} (x) tan^{3} (x) d x

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (x)

1 - 6 x

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