What is the integral from 0 to 1 of (x^2-1)/(1+x^2) ?

The integral from 0 to 1 of (x^2-1)/(1+x^2) is -pi/2+1

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integral from 0 to 1 of (x^2-1)/(1+x^2)

\int_{0}^{1} \frac{x ^{2} - 1}{1 + x ^{2}} d x

- \frac{π}{2} + 1

Solution steps

\int_{0}^{1} \frac{x ^{2} - 1}{1 + x ^{2}} d x

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

= \int_{0}^{1} - \frac{2}{1 + x ^{2}} + 1 d x

Apply the Sum Rule:

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

= - \int_{0}^{1} \frac{2}{1 + x ^{2}} d x + \int_{0}^{1} 1 d x

\int_{0}^{1} \frac{2}{1 + x ^{2}} d x = \frac{π}{2}

\int_{0}^{1} 1 d x = 1

= - \frac{π}{2} + 1

\int_{1}^{2} x^{2} + \frac{6}{2} x d x

\int_{0}^{4} x^{2} - 5 d x

\int_{4}^{16} (y - \frac{y}{4}) d y

\int_{2}^{4} \frac{1}{4} d x

\int_{1}^{4} (x^{2} + 2 x) d x

What is the integral from 0 to 1 of (x^2-1)/(1+x^2) ?
The integral from 0 to 1 of (x^2-1)/(1+x^2) is -pi/2+1