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

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

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integral from-1 to 1 of (x^3+1)/(x^2+1)

\int_{- 1}^{1} \frac{x ^{3} + 1}{x ^{2} + 1} d x

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{- 1}^{1} \frac{x ^{3} + 1}{x ^{2} + 1} d x

Expand

\frac{x ^{3} + 1}{x ^{2} + 1} : \frac{x ^{3}}{x ^{2} + 1} + \frac{1}{x ^{2} + 1}

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}^{1} \frac{x ^{3}}{x ^{2} + 1} d x + \int_{- 1}^{1} \frac{1}{x ^{2} + 1} d x

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

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

= 0 + \frac{π}{2}

Simplify

= \frac{π}{2}

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