What is the integral from 0 to 5 of sqrt(x^2+3x+1) ?

The integral from 0 to 5 of sqrt(x^2+3x+1) is (-5ln(\frac{333+52sqrt(41))/(25))+4(13sqrt(41)-3)}{16}

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integral from 0 to 5 of sqrt(x^2+3x+1)

\int_{0}^{5} x^{2} + 3 x + 1 d x

\frac{- 5 ln ( \frac{333 + 52 41}{25} ) + 4 ( 13 41 - 3 )}{16}

Decimal

19.03441 \dots

Solution steps

\int_{0}^{5} x^{2} + 3 x + 1 d x

Complete the square

x^{2} + 3 x + 1 : (x + \frac{3}{2})^{2} - \frac{5}{4}

= \int_{0}^{5} (x + \frac{3}{2})^{2} - \frac{5}{4} d x

Apply u-substitution

: \int_{\frac{3}{2}}^{\frac{13}{2}} \frac{4 u ^{2} - 5}{2} d u

= \int_{\frac{3}{2}}^{\frac{13}{2}} \frac{4 u ^{2} - 5}{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_{\frac{3}{2}}^{\frac{13}{2}} 4 u^{2} - 5 d u

Apply Trigonometric Substitution

: \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} \frac{5}{2} tan^{2} (v) sec (v) d v

= \frac{1}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} \frac{5}{2} 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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} tan^{2} (v) sec (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} (- 1 + sec^{2} (v)) sec (v) d v

Expand

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

= \frac{1}{2} \cdot \frac{5}{2} \cdot \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} - 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

= \frac{1}{2} \cdot \frac{5}{2} (- \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec (v) d v + \int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec^{3} (v) d v)

\int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec (v) d v = ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)

\int_{arcsec (\frac{3}{5})}^{arcsec (\frac{13}{5})} sec^{3} (v) d v = \frac{13 41 - 3}{5} + \frac{1}{2} (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5))

= \frac{1}{2} \cdot \frac{5}{2} (- (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)) + \frac{13 41 - 3}{5} + \frac{1}{2} (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)))

Simplify

\frac{1}{2} \cdot \frac{5}{2} (- (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5)) + \frac{13 41 - 3}{5} + \frac{1}{2} (ln (\frac{2 205 + 13 5}{5}) - \frac{1}{2} ln (5))) : \frac{- 5 ln ( \frac{333 + 52 41}{25} ) + 4 ( 13 41 - 3 )}{16}

= \frac{- 5 ln ( \frac{333 + 52 41}{25} ) + 4 ( 13 41 - 3 )}{16}

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What is the integral from 0 to 5 of sqrt(x^2+3x+1) ?
The integral from 0 to 5 of sqrt(x^2+3x+1) is (-5ln(\frac{333+52sqrt(41))/(25))+4(13sqrt(41)-3)}{16}