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

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

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integral of sqrt(9+y^2)

\int 9 + y^{2} d y

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

Solution steps

\int 9 + y^{2} d y

Apply Trigonometric Substitution

: \int 9 sec^{3} (u) d u

= \int 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 sec^{3} (u) d u

Apply Integral Reduction

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 9 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{1}{3} y)

= 9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} y ) ) sin ( arctan ( \frac{1}{3} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} y)) + sec (arctan (\frac{1}{3} y)))

Simplify

9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} y ) ) sin ( arctan ( \frac{1}{3} y ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} y)) + sec (arctan (\frac{1}{3} y))) : 9 \frac{1}{18} y 9 + y^{2} + \frac{1}{2} ln \frac{y + 9 + y ^{2}}{3}

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

Add a constant to the solution

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

f (x) = \frac{3}{x ^{2}}, at x = 2

\frac{\partial}{\partial x} (y^{3 x})

f (x) = x + 2, x, x = 1, x = 3

\frac{d}{d x} (\frac{25}{x})

\int \frac{x}{x ^{2} + 5} d x

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