What is the integral of (7xln(x))/(sqrt(x^2-9)) ?

The integral of (7xln(x))/(sqrt(x^2-9)) is 7(ln(x)sqrt(x^2-9)+3arcsec(1/3 x)-sqrt(x^2-9))+C

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integral of (7xln(x))/(sqrt(x^2-9))

\int \frac{7 x ln ( x )}{x ^{2} - 9} d x

7 (ln (x) x^{2} - 9 + 3 arcsec (\frac{1}{3} x) - x^{2} - 9) + C

Solution steps

\int \frac{7 x ln ( x )}{x ^{2} - 9} d x

Take the constant out:

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

= 7 \cdot \int \frac{x ln ( x )}{x ^{2} - 9} d x

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

= 7 \cdot \int x ln (x) \frac{1}{x ^{2} - 9} d x

Apply Integration By Parts

: ln (x) x^{2} - 9 - \int \frac{x ^{2} - 9}{x} d x

= 7 (ln (x) x^{2} - 9 - \int \frac{x ^{2} - 9}{x} d x)

\int \frac{x ^{2} - 9}{x} d x = - 3 arcsec (\frac{1}{3} x) + x^{2} - 9

= 7 (ln (x) x^{2} - 9 - (- 3 arcsec (\frac{1}{3} x) + x^{2} - 9))

Simplify

= 7 (ln (x) x^{2} - 9 + 3 arcsec (\frac{1}{3} x) - x^{2} - 9)

Add a constant to the solution

= 7 (ln (x) x^{2} - 9 + 3 arcsec (\frac{1}{3} x) - x^{2} - 9) + C

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What is the integral of (7xln(x))/(sqrt(x^2-9)) ?
The integral of (7xln(x))/(sqrt(x^2-9)) is 7(ln(x)sqrt(x^2-9)+3arcsec(1/3 x)-sqrt(x^2-9))+C