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

The integral from 0 to 1 of (x^2+1)^2 is 28/15

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

\int_{0}^{1} (x^{2} + 1)^{2} d x

\frac{28}{15}

Decimal Notation

1.86666 \dots

Solution steps

\int_{0}^{1} (x^{2} + 1)^{2} d x

Expand

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

= \int_{0}^{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_{0}^{1} x^{4} d x + \int_{0}^{1} 2 x^{2} d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x^{4} d x = \frac{1}{5}

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

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

= \frac{1}{5} + \frac{2}{3} + 1

Simplify

\frac{1}{5} + \frac{2}{3} + 1 : \frac{28}{15}

= \frac{28}{15}

t \to 6 lim (8 (t - 5) (t - 7))

\frac{d}{d x} (\frac{x + \frac{3}{x}}{x})

\int (x^{2} + 1) d x

\int 2^{5 x} d x

y = x^{2} - 7, (3, 2)

What is the integral from 0 to 1 of (x^2+1)^2 ?
The integral from 0 to 1 of (x^2+1)^2 is 28/15