What is the integral of (4x)/(sqrt(x^2-9)) ?

The integral of (4x)/(sqrt(x^2-9)) is 4sqrt(x^2-9)+C

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integral of (4x)/(sqrt(x^2-9))

\int \frac{4 x}{x ^{2} - 9} d x

4 x^{2} - 9 + C

Solution steps

\int \frac{4 x}{x ^{2} - 9} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{2} - 9} d x

Apply u-substitution

= 4 \cdot \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

= 4 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} - 9

= 4 \cdot \frac{1}{2} \cdot 2 (x^{2} - 9)^{\frac{1}{2}}

Simplify

4 \cdot \frac{1}{2} \cdot 2 (x^{2} - 9)^{\frac{1}{2}} : 4 x^{2} - 9

= 4 x^{2} - 9

Add a constant to the solution

= 4 x^{2} - 9 + C

What is the integral of (4x)/(sqrt(x^2-9)) ?
The integral of (4x)/(sqrt(x^2-9)) is 4sqrt(x^2-9)+C