What is the integral from-1 to 3 of sqrt(1+(2y+2)^2) ?

The integral from-1 to 3 of sqrt(1+(2y+2)^2) is 1/2 (4sqrt(65)+1/2 ln(8+sqrt(65)))

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integral from-1 to 3 of sqrt(1+(2y+2)^2)

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

\frac{1}{2} (465 + \frac{1}{2} ln (8 + 65))

Decimal Notation

16.81863 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{1 + u ^{2}}{2} d u

= \int_{0}^{8} \frac{1 + u ^{2}}{2} 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_{0}^{8} 1 + u^{2} d u

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (8)} sec^{3} (v) d v

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (8)} sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (8)} sec (v) d v)

\int_{0}^{a r c t a n (8)} sec (v) d v = ln (8 + 65)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65)) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{0}^{a r c t a n (8)} + \frac{1}{2} ln (8 + 65))

Compute the boundaries:

465

= \frac{1}{2} (465 + \frac{1}{2} ln (8 + 65))

x^{3}, at x = - 2

\frac{s - 7}{s + 5}

\int tan (7 x se) c^{2} 7 x d x

\int e^{- 2 x} cos (3 x) d x

x \to 3 lim (x^{2} + 16)

What is the integral from-1 to 3 of sqrt(1+(2y+2)^2) ?
The integral from-1 to 3 of sqrt(1+(2y+2)^2) is 1/2 (4sqrt(65)+1/2 ln(8+sqrt(65)))