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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Substitute back

u = arctan (y)

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

Simplify

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

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

Add a constant to the solution

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

\int - 27 e^{x} d x

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\int x (x - 5) \frac{1}{2} d x

\int 3 x^{2} - 3 - 4 e^{2 x} d x

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

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