What is the integral of (5xln(x))/(sqrt(x^2-4)) ?

The integral of (5xln(x))/(sqrt(x^2-4)) is 5(ln(x)sqrt(x^2-4)+2arcsec(1/2 x)-sqrt(x^2-4))+C

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integral of (5xln(x))/(sqrt(x^2-4))

\int \frac{5 x ln ( x )}{x ^{2} - 4} d x

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

Solution steps

\int \frac{5 x ln ( x )}{x ^{2} - 4} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ln ( x )}{x ^{2} - 4} d x

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

= 5 \cdot \int x ln (x) \frac{1}{x ^{2} - 4} d x

Apply Integration By Parts

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

= 5 (ln (x) x^{2} - 4 - \int \frac{x ^{2} - 4}{x} d x)

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

= 5 (ln (x) x^{2} - 4 - (- 2 arcsec (\frac{1}{2} x) + x^{2} - 4))

Simplify

= 5 (ln (x) x^{2} - 4 + 2 arcsec (\frac{1}{2} x) - x^{2} - 4)

Add a constant to the solution

= 5 (ln (x) x^{2} - 4 + 2 arcsec (\frac{1}{2} x) - x^{2} - 4) + C

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What is the integral of (5xln(x))/(sqrt(x^2-4)) ?
The integral of (5xln(x))/(sqrt(x^2-4)) is 5(ln(x)sqrt(x^2-4)+2arcsec(1/2 x)-sqrt(x^2-4))+C