What is the integral from 0 to 1 of (x^2-5)/(1+x^2) ?

The integral from 0 to 1 of (x^2-5)/(1+x^2) is -(3pi)/2+1

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integral from 0 to 1 of (x^2-5)/(1+x^2)

\int_{0}^{1} \frac{x ^{2} - 5}{1 + x ^{2}} d x

- \frac{3 π}{2} + 1

Decimal

- 3.71238 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2} - 5}{1 + x ^{2}} d x

\frac{x ^{2} - 5}{1 + x ^{2}} = - \frac{6}{1 + x ^{2}} + 1

= \int_{0}^{1} - \frac{6}{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{6}{1 + x ^{2}} d x + \int_{0}^{1} 1 d x

\int_{0}^{1} \frac{6}{1 + x ^{2}} d x = \frac{3 π}{2}

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

= - \frac{3 π}{2} + 1

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What is the integral from 0 to 1 of (x^2-5)/(1+x^2) ?
The integral from 0 to 1 of (x^2-5)/(1+x^2) is -(3pi)/2+1