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

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

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

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

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

Solution steps

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

(- 2.7) (2.3)

\int_{0}^{1} x e^{2 x^{2}} d x

n = 1 \sum \infty p (1 - p)^{2 n}

\frac{d y}{d x} = \frac{y ^{2} + 1}{x ^{2} + 1}, y (0) = 1

\int \frac{3 x ^{7}}{4 + x ^{8}} d x

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