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

The integral from-1 to 3 of sqrt(1+(2y+1)^2) is 1/2 (18sqrt(2)+1/2 (ln(7+5sqrt(2))-ln(-1+sqrt(2))))

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

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

\frac{1}{2} (182 + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

Decimal Notation

13.60929 \dots

Solution steps

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

Apply u-substitution

: \int_{- 1}^{7} \frac{1 + u ^{2}}{2} d u

= \int_{- 1}^{7} \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_{- 1}^{7} 1 + u^{2} d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{4}}^{a r c t a n (7)} sec^{3} (v) d v

= \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{a r c t a n (7)} sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} \cdot \int_{- \frac{π}{4}}^{a r c t a n (7)} sec (v) d v)

\int_{- \frac{π}{4}}^{a r c t a n (7)} sec (v) d v = ln (7 + 52) - ln (- 1 + 2)

= \frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

Simplify

\frac{1}{2} ([\frac{sec ^{2} ( v ) sin ( v )}{2}]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2))) : \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

= \frac{1}{2} ([\frac{1}{2} sec (v) tan (v)]_{- \frac{π}{4}}^{a r c t a n (7)} + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

Compute the boundaries:

182

= \frac{1}{2} (182 + \frac{1}{2} (ln (7 + 52) - ln (- 1 + 2)))

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What is the integral from-1 to 3 of sqrt(1+(2y+1)^2) ?
The integral from-1 to 3 of sqrt(1+(2y+1)^2) is 1/2 (18sqrt(2)+1/2 (ln(7+5sqrt(2))-ln(-1+sqrt(2))))