What is the integral from 4 to 6 of 1/(sqrt(t^2-9)) ?

The integral from 4 to 6 of 1/(sqrt(t^2-9)) is ln(-((2+sqrt(3))(sqrt(7)-4))/3)

Popular >

integral from 4 to 6 of 1/(sqrt(t^2-9))

\int_{4}^{6} \frac{1}{t ^{2} - 9} d t

ln (- \frac{( 2 + 3 ) ( 7 - 4 )}{3})

Decimal Notation

0.52159 \dots

Solution steps

\int_{4}^{6} \frac{1}{t ^{2} - 9} d t

Apply Trigonometric Substitution

: \int_{arcsec (\frac{4}{3})}^{\frac{π}{3}} sec (u) d u

= \int_{arcsec (\frac{4}{3})}^{\frac{π}{3}} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= [ln ∣ tan (u) + sec (u) ∣]_{arcsec (\frac{4}{3})}^{\frac{π}{3}}

Compute the boundaries:

ln (3 + 2) - ln (\frac{7}{3} + \frac{4}{3})

= ln (3 + 2) - ln (\frac{7}{3} + \frac{4}{3})

Simplify

= ln (- \frac{( 2 + 3 ) ( 7 - 4 )}{3})

\frac{d}{d x} (\frac{1}{1 + y ^{2}} + cos (x) - 2 x y)

4 e^{x - 3}

\frac{d}{d y} (x^{2} + y^{2} + z^{2})

x \cdot \frac{d y}{d x} + y = 4 x + 1

\frac{\partial}{\partial x} (y arctan (\frac{x}{y}))

What is the integral from 4 to 6 of 1/(sqrt(t^2-9)) ?
The integral from 4 to 6 of 1/(sqrt(t^2-9)) is ln(-((2+sqrt(3))(sqrt(7)-4))/3)