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

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

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

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

2 arctan (t) - t + C

Solution steps

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

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

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

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

\int 1 d t = t

= 2 arctan (t) - t

Add a constant to the solution

= 2 arctan (t) - t + C

\int \frac{1}{( y ^{2} - y )} d y

x^{2} y^{^{''}} = (y^{^{'}})^{2}

\frac{\partial}{\partial x} (3 y^{3} - 4 y x^{2})

\int_{0}^{7} f (x) d x

θ \to 0 lim (\frac{sin ( 4 θ )}{θ})

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