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

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

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

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

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

Solution steps

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

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 (x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x + 10)

\int (x \cdot e^{x}) d x

\frac{d}{d x} (y z e^{x z})

\frac{d}{d x} (arcsin (4 x^{5}))

\frac{d y}{d x}, 9 sin (x) + cos (y) = sin (x) cos (y)

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