What is the integral from 3 to 6 of sqrt(1+x^4+2x^2) ?

The integral from 3 to 6 of sqrt(1+x^4+2x^2) is 66

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integral from 3 to 6 of sqrt(1+x^4+2x^2)

\int_{3}^{6} 1 + x^{4} + 2 x^{2} d x

66

Solution steps

\int_{3}^{6} 1 + x^{4} + 2 x^{2} d x

Factor

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

= \int_{3}^{6} (x^{2} + 1)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{3}^{6} (x^{2} + 1)^{2} d x

Eliminate Absolutes

= \int_{3}^{6} (x^{2} + 1)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{3}^{6} (x^{2} + 1) d x

Eliminate Absolutes

= \int_{3}^{6} 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_{3}^{6} x^{2} d x + \int_{3}^{6} 1 d x

\int_{3}^{6} x^{2} d x = 63

\int_{3}^{6} 1 d x = 3

= 63 + 3

Simplify

= 66

What is the integral from 3 to 6 of sqrt(1+x^4+2x^2) ?
The integral from 3 to 6 of sqrt(1+x^4+2x^2) is 66