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

The integral of ln(sqrt(x^2-4)) is 1/2 (xln(x^2-4)+4(1/2 ln| x/2+1|-1/2 ln| x/2-1|-x/2))+C

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

\int ln (x^{2} - 4) d x

\frac{1}{2} (x ln (x^{2} - 4) + 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})) + C

Solution steps

\int ln (x^{2} - 4) d x

Simplify

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

= \int \frac{1}{2} ln (x^{2} - 4) d x

Take the constant out:

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

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

Apply Integration By Parts

: x ln (x^{2} - 4) - \int \frac{2 x ^{2}}{x ^{2} - 4} d x

= \frac{1}{2} (x ln (x^{2} - 4) - \int \frac{2 x ^{2}}{x ^{2} - 4} d x)

\int \frac{2 x ^{2}}{x ^{2} - 4} d x = - 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})

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

Simplify

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

Add a constant to the solution

= \frac{1}{2} (x ln (x^{2} - 4) + 4 (\frac{1}{2} ln \frac{x}{2} + 1 - \frac{1}{2} ln \frac{x}{2} - 1 - \frac{x}{2})) + C

n = 0 \sum \infty \frac{n ^{7}}{7 ^{n}}

\frac{d}{d x} ((4 x^{2} + x) (x - x^{2}))

\int_{0}^{3} 5 x - x^{2} - 2 x d x

\frac{1}{9 x ^{\frac{8}{9}}}

\int x csc (4) x^{2} d x

What is the integral of ln(sqrt(x^2-4)) ?
The integral of ln(sqrt(x^2-4)) is 1/2 (xln(x^2-4)+4(1/2 ln| x/2+1|-1/2 ln| x/2-1|-x/2))+C