What is the integral of (-4)/((x^2+4)^2) ?

The integral of (-4)/((x^2+4)^2) is -1/4 (arctan(x/2)+1/2 sin(2arctan(x/2)))+C

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integral of (-4)/((x^2+4)^2)

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

- \frac{1}{4} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) + C

Solution steps

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

Take the constant out:

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

= - 4 \cdot \int \frac{1}{( x ^{2} + 4 ) ^{2}} d x

Apply Integral Substitution

= - 4 \cdot \int \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{8} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply Trigonometric Substitution

= - 4 \cdot \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Rewrite using trig identities

= - 4 \cdot \frac{1}{8} \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

= - 4 \cdot \frac{1}{8} \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

= - 4 \cdot \frac{1}{8} \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)

= - 4 \cdot \frac{1}{8} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= - 4 \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

Simplify

- 4 \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : - \frac{1}{4} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

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

Add a constant to the solution

= - \frac{1}{4} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) + C

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