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))

integral

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

- 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

\frac{e ^{4} ln ( x ) - e ^{5} ln ( x )}{x ^{5}}

\frac{1 - x ^{4}}{1 - x}

\frac{( cos ^{3} ( x ) )}{sin ^{4} ( x )}

(t^{2})^{2}

\frac{1}{( 2 x - x ^{2} )}

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