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

The integral from 0 to y of sqrt(9+y^2) is 9((ysqrt(y^2+9))/(18)+1/2 ln(1/3 |y+sqrt(y^2+9)|))

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

\int_{0}^{y} 9 + y^{2} d y

9 (\frac{y y ^{2} + 9}{18} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

Solution steps

\int_{0}^{y} 9 + y^{2} d y

Apply Trigonometric Substitution

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

= \int_{0}^{a r c t a n (\frac{y}{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{y}{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{y}{3})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{y}{3})} sec (u) d u

\int_{0}^{a r c t a n (\frac{y}{3})} sec (u) d u = ln (\frac{1}{3} y + y^{2} + 9)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{18} y y^{2} + 9

= 9 (\frac{1}{18} y y^{2} + 9 + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

Simplify

= 9 (\frac{y y ^{2} + 9}{18} + \frac{1}{2} ln (\frac{1}{3} y + y^{2} + 9))

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What is the integral from 0 to y of sqrt(9+y^2) ?
The integral from 0 to y of sqrt(9+y^2) is 9((ysqrt(y^2+9))/(18)+1/2 ln(1/3 |y+sqrt(y^2+9)|))