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

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

= \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

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