What is the integral from 1 to 4 of sqrt(x^2+3) ?

The integral from 1 to 4 of sqrt(x^2+3) is 2sqrt(19)-1+3/2 ln(4/(sqrt(3))+sqrt(19/3))-3/4 ln(3)

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integral from 1 to 4 of sqrt(x^2+3)

\int_{1}^{4} x^{2} + 3 d x

219 - 1 + \frac{3}{2} ln (\frac{4}{3} + \frac{19}{3}) - \frac{3}{4} ln (3)

Decimal Notation

9.25486 \dots

Solution steps

\int_{1}^{4} x^{2} + 3 d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} 3 sec^{3} (u) d u

= \int_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} 3 sec^{3} (u) d u

Take the constant out:

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

= 3 \cdot \int_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} sec^{3} (u) d u

Apply Integral Reduction

= 3 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} \cdot \int_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} sec (u) d u

\int_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} sec (u) d u = ln (\frac{4}{3} + \frac{19}{3}) - \frac{1}{2} ln (3)

= 3 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} (ln (\frac{4}{3} + \frac{19}{3}) - \frac{1}{2} ln (3))

Simplify

3 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} (ln (\frac{4}{3} + \frac{19}{3}) - \frac{1}{2} ln (3)) : 3 [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} + \frac{3}{2} ln (\frac{4}{3} + \frac{19}{3}) - \frac{3}{4} ln (3)

= 3 [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{6}}^{a r c t a n (\frac{4}{3})} + \frac{3}{2} ln (\frac{4}{3} + \frac{19}{3}) - \frac{3}{4} ln (3)

Compute the boundaries:

\frac{2 19}{3} - \frac{1}{3}

= 3 (\frac{2 19}{3} - \frac{1}{3}) + \frac{3}{2} ln (\frac{4}{3} + \frac{19}{3}) - \frac{3}{4} ln (3)

Simplify

= 219 - 1 + \frac{3}{2} ln (\frac{4}{3} + \frac{19}{3}) - \frac{3}{4} ln (3)

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What is the integral from 1 to 4 of sqrt(x^2+3) ?
The integral from 1 to 4 of sqrt(x^2+3) is 2sqrt(19)-1+3/2 ln(4/(sqrt(3))+sqrt(19/3))-3/4 ln(3)