What is the integral of 9xarcsec(x) ?

The integral of 9xarcsec(x) is 9(1/2 x^2arcsec(x)-1/2 sqrt(x^2-1))+C

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integral of 9xarcsec(x)

\int 9 x arcsec (x) d x

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

Solution steps

\int 9 x arcsec (x) d x

Take the constant out:

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

= 9 \cdot \int x arcsec (x) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} arcsec (x) - \int \frac{x ^{2}}{2 x ^{2} x ^{2} - 1} d x

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

\int \frac{x ^{2}}{2 x ^{2} x ^{2} - 1} d x = \frac{1}{2} x^{2} - 1

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

Add a constant to the solution

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

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