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

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

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

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

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

= 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

f (x) = x^{2} + 7, (- 4, 23)

x \to 1 + lim (x^{\frac{4}{1 - x}})

\int \frac{- 2 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

x \to 1 + lim (\frac{4}{x - 1})

\frac{\partial}{\partial x} (2 x^{2} y - 2 x + 4 y^{3})

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