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

The integral from-x^2 to x of (t^2)/(1+t^2) is -arctan(x)+arctan(-x^2)+x+x^2

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integral from-x^2 to x of (t^2)/(1+t^2)

\int_{- x^{2}}^{x} \frac{t ^{2}}{1 + t ^{2}} d t

- arctan (x) + arctan (- x^{2}) + x + x^{2}

Solution steps

\int_{- x^{2}}^{x} \frac{t ^{2}}{1 + t ^{2}} d t

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

= \int_{- x^{2}}^{x} - \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_{- x^{2}}^{x} \frac{1}{1 + t ^{2}} d t + \int_{- x^{2}}^{x} 1 d t

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

\int_{- x^{2}}^{x} 1 d t = x + x^{2}

= - (arctan (x) - arctan (- x^{2})) + x + x^{2}

Simplify

= - arctan (x) + arctan (- x^{2}) + x + x^{2}

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