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

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

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

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

2 x^{2} - 4 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 4

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

Simplify

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

= 2 x^{2} - 4

Add a constant to the solution

= 2 x^{2} - 4 + C

\int 2 e^{- 2 x} sin (2 x) d x

\frac{d y}{d x} - 2 y tan (x) = y^{2} tan^{2} (x)

\int \frac{x ^{2}}{x - 6} d x

f (x) = \frac{x ^{2}}{e ^{x}}

y = \frac{2}{ln ( x )}

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