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

The integral from-2 to 2 of sqrt(1+(2x)^2) is 1/2 (4sqrt(17)+1/2 (ln(4+sqrt(17))-ln(-4+sqrt(17))))

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

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

\frac{1}{2} (417 + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

Decimal Notation

9.29356 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{- a r c t a n (4)}^{a r c t a n (4)} \frac{1}{2} sec^{3} (u) d u

= \int_{- a r c t a n (4)}^{a r c t a n (4)} \frac{1}{2} sec^{3} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{- a r c t a n (4)}^{a r c t a n (4)} sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (4)}^{a r c t a n (4)} + \frac{1}{2} \cdot \int_{- a r c t a n (4)}^{a r c t a n (4)} sec (u) d u)

\int_{- a r c t a n (4)}^{a r c t a n (4)} sec (u) d u = ln (4 + 17) - ln (- 4 + 17)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (4)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{- a r c t a n (4)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17))) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (4)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{- a r c t a n (4)}^{a r c t a n (4)} + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

Compute the boundaries:

417

= \frac{1}{2} (417 + \frac{1}{2} (ln (4 + 17) - ln (- 4 + 17)))

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What is the integral from-2 to 2 of sqrt(1+(2x)^2) ?
The integral from-2 to 2 of sqrt(1+(2x)^2) is 1/2 (4sqrt(17)+1/2 (ln(4+sqrt(17))-ln(-4+sqrt(17))))