What is the integral of 1/(sqrt((16+x^2)^4)) ?

The integral of 1/(sqrt((16+x^2)^4)) is 1/128 (arctan(x/4)+(4x)/(16+x^2))+C

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integral of 1/(sqrt((16+x^2)^4))

\int \frac{1}{( 16 + x ^{2} ) ^{4}} d x

\frac{1}{128} (arctan (\frac{x}{4}) + \frac{4 x}{16 + x ^{2}}) + C

Solution steps

\int \frac{1}{( 16 + x ^{2} ) ^{4}} d x

(16 + x^{2})^{4} = (16 + x^{2})^{2},

assuming

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

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

Apply Integral Substitution

= \int \frac{1}{64 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= \frac{1}{64} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{64} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{64} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{64} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{64} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= \frac{1}{64} \cdot \frac{1}{2} (arctan (\frac{x}{4}) + \frac{1}{2} sin (2 arctan (\frac{x}{4})))

Simplify

\frac{1}{64} \cdot \frac{1}{2} (arctan (\frac{x}{4}) + \frac{1}{2} sin (2 arctan (\frac{x}{4}))) : \frac{1}{128} (arctan (\frac{x}{4}) + \frac{4 x}{16 + x ^{2}})

= \frac{1}{128} (arctan (\frac{x}{4}) + \frac{4 x}{16 + x ^{2}})

Add a constant to the solution

= \frac{1}{128} (arctan (\frac{x}{4}) + \frac{4 x}{16 + x ^{2}}) + C

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What is the integral of 1/(sqrt((16+x^2)^4)) ?
The integral of 1/(sqrt((16+x^2)^4)) is 1/128 (arctan(x/4)+(4x)/(16+x^2))+C