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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Substitute back

v = arctan (u)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{4} sin (4 x) + c)

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

\int \frac{sin ( 2 x )}{16 + cos ^{2} ( x )} d x

6 x y^{9} + 5 x y = 11, (1, 1)

x \to + (- 2) lim (x - 2)

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