What is the integral from 1 to 8 of 1/(xsqrt(16x^2-9)) ?

The integral from 1 to 8 of 1/(xsqrt(16x^2-9)) is 1/3 (arcsec(32/3)-arcsec(4/3))

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integral from 1 to 8 of 1/(xsqrt(16x^2-9))

\int_{1}^{8} \frac{1}{x 16 x ^{2} - 9} d x

\frac{1}{3} (arcsec (\frac{32}{3}) - arcsec (\frac{4}{3}))

Decimal Notation

0.25139 \dots

Solution steps

\int_{1}^{8} \frac{1}{x 16 x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} \frac{1}{3} d u

= \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})} 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} [1 \cdot u]_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})}

Simplify

= \frac{1}{3} [u]_{arcsec (\frac{4}{3})}^{arcsec (\frac{32}{3})}

Compute the boundaries:

arcsec (\frac{32}{3}) - arcsec (\frac{4}{3})

= \frac{1}{3} (arcsec (\frac{32}{3}) - arcsec (\frac{4}{3}))

x \to 6 lim (cos (\frac{1}{x - 6}))

\int cos (x) sec^{3} (x) d x

f (x) = \frac{x ^{3} + 3 x + 2}{x ^{2} - 1}

\int_{0}^{2} \int_{0}^{2 - x^{2}} 18 x y d y d x

x \to - 2 lim (\frac{x + 2}{2 x + x ^{2} + 12})

What is the integral from 1 to 8 of 1/(xsqrt(16x^2-9)) ?
The integral from 1 to 8 of 1/(xsqrt(16x^2-9)) is 1/3 (arcsec(32/3)-arcsec(4/3))