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

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

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integral of sqrt(x^2+9)

\int x^{2} + 9 d x

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

Solution steps

\int x^{2} + 9 d x

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} x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} - 2 t y = 9 t^{2} e^{t^{2}}

\frac{\partial}{\partial x} (x sin (x^{2} y))

y^{8} - \frac{x ^{4} y ^{9} + 4}{x ^{5} + 4 y ^{^{'}}} = 0

\int_{0}^{1} 1 + 4 x^{2} d x

\frac{3 x - 2}{2 x + 1}

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