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

The integral of x/(sqrt(x^2-9)) is sqrt(x^2-9)+C

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integral of x/(sqrt(x^2-9))

\int \frac{x}{x ^{2} - 9} d x

x^{2} - 9 + C

Solution steps

\int \frac{x}{x ^{2} - 9} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 9

= \frac{1}{2} \cdot 2 (x^{2} - 9)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (x^{2} - 9)^{\frac{1}{2}} : x^{2} - 9

= x^{2} - 9

Add a constant to the solution

= x^{2} - 9 + C

\frac{d}{d x} (3^{x^{5} + 2})

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

x v^{'} + 2 v = 0

x y^{^{'}} = 2 y + x^{3} cos (x)

\int cos^{5} (x) sin^{4} (x) d x

What is the integral of x/(sqrt(x^2-9)) ?
The integral of x/(sqrt(x^2-9)) is sqrt(x^2-9)+C