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

The integral from-1 to 1 of (x^2+1)^2 is 56/15

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

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

\frac{56}{15}

Decimal Notation

3.73333 \dots

Solution steps

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

Expand

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

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

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

\int_{- 1}^{1} 2 x^{2} d x = \frac{4}{3}

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

= \frac{2}{5} + \frac{4}{3} + 2

Simplify

\frac{2}{5} + \frac{4}{3} + 2 : \frac{56}{15}

= \frac{56}{15}

y = e^{- \frac{1}{2} x}

(2 x^{3} + 5 x) (x - 3) (x + 2)

\frac{d}{d x} (ln (y - x))

- x^{2} + 8 x - 9, (\frac{1}{2} x) + 1, x = 3, x = 5

\int_{- \infty}^{0} \frac{1}{7 - 5 x} d x

What is the integral from-1 to 1 of (x^2+1)^2 ?
The integral from-1 to 1 of (x^2+1)^2 is 56/15