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

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

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

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

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

Solution steps

\int \frac{1}{( X ^{2} + 1 ) ^{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

x \to + 0 - lim (\frac{2}{x})

\int csc^{2} (\frac{x}{8}) d x

\int \frac{1}{x - x} d x

\frac{d y}{d x} + (\frac{3}{x}) y = - 3 x - 5

\int_{- 1}^{2} t - 6 t^{2} d t

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