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

The integral from-1 to 0 of (x^2-x)/(1+x^2) is -pi/4+1+1/2 ln(2)

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

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

- \frac{π}{4} + 1 + \frac{1}{2} ln (2)

Decimal Notation

0.56117 \dots

Solution steps

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

Expand

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

Apply the Sum Rule:

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

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

\int_{- 1}^{0} \frac{x ^{2}}{1 + x ^{2}} d x = - \frac{π}{4} + 1

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

= - \frac{π}{4} + 1 - (- \frac{1}{2} ln (2))

Simplify

= - \frac{π}{4} + 1 + \frac{1}{2} ln (2)

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What is the integral from-1 to 0 of (x^2-x)/(1+x^2) ?
The integral from-1 to 0 of (x^2-x)/(1+x^2) is -pi/4+1+1/2 ln(2)