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

The integral from 0 to 3 of sqrt(1+x^4+2x^2) is 12

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

\int_{0}^{3} 1 + x^{4} + 2 x^{2} d x

12

Solution steps

\int_{0}^{3} 1 + x^{4} + 2 x^{2} d x

Factor

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

= \int_{0}^{3} (x^{2} + 1)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{3} (x^{2} + 1)^{2} d x

Eliminate Absolutes

= \int_{0}^{3} (x^{2} + 1)^{2} d x

Apply exponent rule:

a^{2} = ∣ a ∣

= \int_{0}^{3} (x^{2} + 1) d x

Eliminate Absolutes

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

\int_{0}^{3} x^{2} d x = 9

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

= 9 + 3

Simplify

= 12

(x + 4)^{^{'}}

\int \frac{e ^{x}}{1 + 1 + e ^{x}} d x

\frac{d}{d x} (e^{- x} cos (x) + e^{- x} sin (x))

ln (3 x + 5)

f (x) = x^{2} - 4 x, at x = - 2

What is the integral from 0 to 3 of sqrt(1+x^4+2x^2) ?
The integral from 0 to 3 of sqrt(1+x^4+2x^2) is 12