What is the integral from 0 to 1 of sqrt(1+x^4+2x^2) ?

The integral from 0 to 1 of sqrt(1+x^4+2x^2) is 4/3

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integral from 0 to 1 of sqrt(1+x^4+2x^2)

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Factor

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

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

Apply exponent rule:

a^{2} = ∣ a ∣

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

Eliminate Absolutes

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

Apply exponent rule:

a^{2} = ∣ a ∣

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

Eliminate Absolutes

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

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

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

= \frac{1}{3} + 1

Simplify

\frac{1}{3} + 1 : \frac{4}{3}

= \frac{4}{3}

\int \frac{1}{( x + 1 ) ( x - 2 )} d x

\int_{6}^{8} (\frac{x ^{2} + 6}{x ^{2}}) d x

\int \frac{x ^{3} - x ^{2} + x - 1}{x ^{2} + 9} d x

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

x \to 2 lim (lo g_{4} (x - 2))

What is the integral from 0 to 1 of sqrt(1+x^4+2x^2) ?
The integral from 0 to 1 of sqrt(1+x^4+2x^2) is 4/3