What is the integral from 1 to 2 of xarcsec(x) ?

The integral from 1 to 2 of xarcsec(x) is (4pi-3sqrt(3))/6

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integral from 1 to 2 of xarcsec(x)

\int_{1}^{2} x arcsec (x) d x

\frac{4 π - 3 3}{6}

Solution steps

\int_{1}^{2} 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]_{1}^{2}

= [\frac{1}{2} x^{2} arcsec (x) - \int \frac{x ^{2}}{2 x ^{2} x ^{2} - 1} d x]_{1}^{2}

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

= [\frac{1}{2} x^{2} arcsec (x) - \frac{1}{2} x^{2} - 1]_{1}^{2}

Simplify

[\frac{1}{2} x^{2} arcsec (x) - \frac{1}{2} x^{2} - 1]_{1}^{2} : [\frac{1}{2} (x^{2} arcsec (x) - x^{2} - 1)]_{1}^{2}

= [\frac{1}{2} (x^{2} arcsec (x) - x^{2} - 1)]_{1}^{2}

Compute the boundaries:

\frac{4 π - 3 3}{6}

= \frac{4 π - 3 3}{6}

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What is the integral from 1 to 2 of xarcsec(x) ?
The integral from 1 to 2 of xarcsec(x) is (4pi-3sqrt(3))/6