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

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

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

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

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

Solution steps

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

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 (t)

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x}{x - 2} d x

\frac{d}{d x} (\frac{3}{2 x - 8})

\int 1 - 2 sin (x) d x

\int (\frac{1}{2 x ^{3}} - \frac{1}{x ^{4}}) d x

\frac{\partial}{\partial x} (ln (x y^{2}))

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