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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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