What is the integral from 0 to 4 of sqrt(x^2+9) ?

The integral from 0 to 4 of sqrt(x^2+9) is 10+9/2 ln(3)

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integral from 0 to 4 of sqrt(x^2+9)

\int_{0}^{4} x^{2} + 9 d x

10 + \frac{9}{2} ln (3)

Solution steps

\int_{0}^{4} x^{2} + 9 d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{4}{3})} 9 sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{4}{3})} 9 sec^{3} (u) d u

Take the constant out:

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

= 9 \cdot \int_{0}^{a r c t a n (\frac{4}{3})} sec^{3} (u) d u

Apply Integral Reduction

= 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{4}{3})} sec (u) d u

\int_{0}^{a r c t a n (\frac{4}{3})} sec (u) d u = ln (3)

= 9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3)

Simplify

9 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3) : 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3))

= 9 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{4}{3})} + \frac{1}{2} ln (3))

Compute the boundaries:

\frac{10}{9}

= 9 (\frac{10}{9} + \frac{1}{2} ln (3))

Simplify

= 10 + \frac{9}{2} ln (3)

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What is the integral from 0 to 4 of sqrt(x^2+9) ?
The integral from 0 to 4 of sqrt(x^2+9) is 10+9/2 ln(3)