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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

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

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

\int 1 d u = u

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

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

Substitute back

u = arctan (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d ^{2} y}{d t ^{2}} + 10 \frac{d y}{d t} + 25 y = 0

\frac{d}{d x} (\frac{x ^{3} y + x y ^{3}}{x ^{2} + y ^{2}})

f (x) = x - 4, (13, 3)

f (x) = \frac{5}{x + 1}, at x = \frac{- 1}{3}

x \to 0 lim (\frac{0 + 5}{3 ( 0 )})

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