What is the integral from 4 to 6 of 1/(xsqrt(x^2-4)) ?

The integral from 4 to 6 of 1/(xsqrt(x^2-4)) is 1/2 (arcsec(3)-pi/3)

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integral from 4 to 6 of 1/(xsqrt(x^2-4))

\int_{4}^{6} \frac{1}{x x ^{2} - 4} d x

\frac{1}{2} (arcsec (3) - \frac{π}{3})

Decimal Notation

0.09188 \dots

Solution steps

\int_{4}^{6} \frac{1}{x x ^{2} - 4} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{3}}^{arcsec (3)} \frac{1}{2} d u

= \int_{\frac{π}{3}}^{arcsec (3)} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{\frac{π}{3}}^{arcsec (3)} 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} [1 \cdot u]_{\frac{π}{3}}^{arcsec (3)}

Simplify

= \frac{1}{2} [u]_{\frac{π}{3}}^{arcsec (3)}

Compute the boundaries:

arcsec (3) - \frac{π}{3}

= \frac{1}{2} (arcsec (3) - \frac{π}{3})

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\int_{- 7}^{0} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

\int_{0}^{1} \frac{1}{2} e^{- \frac{1}{2} x} d x

\int_{0}^{π} cos^{2} (x) d x

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