What is the integral of 1/(tsqrt(t^2-4)) ?

The integral of 1/(tsqrt(t^2-4)) is 1/2 arcsec(1/2 t)+C

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integral of 1/(tsqrt(t^2-4))

\int \frac{1}{t t ^{2} - 4} d t

\frac{1}{2} arcsec (\frac{1}{2} t) + C

Solution steps

\int \frac{1}{t t ^{2} - 4} d t

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \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 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{2} t)

= \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} t)

Simplify

= \frac{1}{2} arcsec (\frac{1}{2} t)

Add a constant to the solution

= \frac{1}{2} arcsec (\frac{1}{2} t) + C

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What is the integral of 1/(tsqrt(t^2-4)) ?
The integral of 1/(tsqrt(t^2-4)) is 1/2 arcsec(1/2 t)+C