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

The integral from 0 to 2 of sqrt(1+4t+t^2) is -3/2 ln(sqrt(13/3)+4/(sqrt(3)))+3/4 ln(3)+2sqrt(13)-1

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

\int_{0}^{2} 1 + 4 t + t^{2} d t

- \frac{3}{2} ln (\frac{13}{3} + \frac{4}{3}) + \frac{3}{4} ln (3) + 213 - 1

Decimal

4.81570 \dots

Solution steps

\int_{0}^{2} 1 + 4 t + t^{2} d t

Complete the square

1 + 4 t + t^{2} : (t + 2)^{2} - 3

= \int_{0}^{2} (t + 2)^{2} - 3 d t

Apply u-substitution

: \int_{2}^{4} u^{2} - 3 d u

= \int_{2}^{4} u^{2} - 3 d u

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} 3 tan^{2} (v) sec (v) d v

= \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} 3 tan^{2} (v) sec (v) d v

Take the constant out:

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

= 3 \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} tan^{2} (v) sec (v) d v

Rewrite using trig identities

= 3 \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} (- 1 + sec^{2} (v)) sec (v) d v

Expand

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= 3 \cdot \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} - sec (v) + sec^{3} (v) d v

Apply the Sum Rule:

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

= 3 (- \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec (v) d v + \int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec^{3} (v) d v)

\int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec (v) d v = ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)

\int_{\frac{π}{6}}^{arcsec (\frac{4}{3})} sec^{3} (v) d v = \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3))

= 3 (- (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)) + \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)))

Simplify

3 (- (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3)) + \frac{2 13 - 1}{3} + \frac{1}{2} (ln (\frac{13}{3} + \frac{4}{3}) - \frac{1}{2} ln (3))) : - \frac{3}{2} ln (\frac{13}{3} + \frac{4}{3}) + \frac{3}{4} ln (3) + 213 - 1

= - \frac{3}{2} ln (\frac{13}{3} + \frac{4}{3}) + \frac{3}{4} ln (3) + 213 - 1

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What is the integral from 0 to 2 of sqrt(1+4t+t^2) ?
The integral from 0 to 2 of sqrt(1+4t+t^2) is -3/2 ln(sqrt(13/3)+4/(sqrt(3)))+3/4 ln(3)+2sqrt(13)-1