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

The integral from-2 to 1 of (x^2+1)^2 is 78/5

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

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

\frac{78}{5}

Decimal Notation

15.6

Solution steps

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

Expand

(x^{2} + 1)^{2} : x^{4} + 2 x^{2} + 1

= \int_{- 2}^{1} x^{4} + 2 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_{- 2}^{1} x^{4} d x + \int_{- 2}^{1} 2 x^{2} d x + \int_{- 2}^{1} 1 d x

\int_{- 2}^{1} x^{4} d x = \frac{33}{5}

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

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

= \frac{33}{5} + 6 + 3

Simplify

\frac{33}{5} + 6 + 3 : \frac{78}{5}

= \frac{78}{5}

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What is the integral from-2 to 1 of (x^2+1)^2 ?
The integral from-2 to 1 of (x^2+1)^2 is 78/5