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

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

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

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

- arctan (t) + t + C

Solution steps

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

\frac{t ^{2}}{1 + t ^{2}} = - \frac{1}{1 + t ^{2}} + 1

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

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

\int 1 d t = t

= - arctan (t) + t

Add a constant to the solution

= - arctan (t) + t + C

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