What is the integral from 1 to 3 of sqrt(1+(2x\sqrt{x^2-1))^2} ?

The integral from 1 to 3 of sqrt(1+(2x\sqrt{x^2-1))^2} is (52+12sqrt(2))/3-4sqrt(2)-2

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integral from 1 to 3 of sqrt(1+(2x\sqrt{x^2-1))^2}

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

\frac{52 + 12 2}{3} - 42 - 2

Decimal

15.33333 \dots

Solution steps

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

Factor

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

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

Apply u-substitution

: \int_{2 + 1}^{2 \cdot 3 + 1} \frac{u ( u - 1 )}{2} - \frac{u}{2} d u

= \int_{2 + 1}^{2 \cdot 3 + 1} \frac{u ( u - 1 )}{2} - \frac{u}{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2 + 1}^{2 \cdot 3 + 1} \frac{u ( u - 1 )}{2} d u - \int_{2 + 1}^{2 \cdot 3 + 1} \frac{u}{2} d u

\int_{2 + 1}^{2 \cdot 3 + 1} \frac{u ( u - 1 )}{2} d u = \frac{52 + 12 2}{3}

\int_{2 + 1}^{2 \cdot 3 + 1} \frac{u}{2} d u = 42 + 2

= \frac{52 + 12 2}{3} - (42 + 2)

Simplify

= \frac{52 + 12 2}{3} - 42 - 2

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What is the integral from 1 to 3 of sqrt(1+(2x\sqrt{x^2-1))^2} ?
The integral from 1 to 3 of sqrt(1+(2x\sqrt{x^2-1))^2} is (52+12sqrt(2))/3-4sqrt(2)-2