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

The integral of (x^2+4)^{-2} is 1/16 (arctan(x/2)+1/2 sin(2arctan(x/2)))+C

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integral of (x^2+4)^{-2}

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

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

Solution steps

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

Apply Integral Substitution

= \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

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Substitute back

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

Simplify

\frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) : \frac{1}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))

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

Add a constant to the solution

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

\int \frac{1}{1 + z ^{2}} d z

t^{2} y^{^{''}} + 13 t y^{^{'}} + 36 y = 0

9 y^{^{''}} - 60 y^{^{'}} + 51 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

y = x^{2} - 1, (- 2, 3)

What is the integral of (x^2+4)^{-2} ?
The integral of (x^2+4)^{-2} is 1/16 (arctan(x/2)+1/2 sin(2arctan(x/2)))+C