What is the integral of 1/(sqrt(t^2-6t+13)) ?

The integral of 1/(sqrt(t^2-6t+13)) is ln(1/2 |t-3+sqrt(t^2-6t+13)|)+C

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integral of 1/(sqrt(t^2-6t+13))

\int \frac{1}{t ^{2} - 6 t + 13} d t

ln (\frac{1}{2} t - 3 + t^{2} - 6 t + 13) + C

Solution steps

\int \frac{1}{t ^{2} - 6 t + 13} d t

Complete the square

t^{2} - 6 t + 13 : (t - 3)^{2} + 4

= \int \frac{1}{( t - 3 ) ^{2} + 4} d t

Apply u-substitution

= \int \frac{1}{u ^{2} + 4} d u

Apply Trigonometric Substitution

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} (t - 3))) + sec (arctan (\frac{1}{2} (t - 3)))

Simplify

ln tan (arctan (\frac{1}{2} (t - 3))) + sec (arctan (\frac{1}{2} (t - 3))) : ln (\frac{1}{2} t - 3 + t^{2} - 6 t + 13)

= ln (\frac{1}{2} t - 3 + t^{2} - 6 t + 13)

Add a constant to the solution

= ln (\frac{1}{2} t - 3 + t^{2} - 6 t + 13) + C

What is the integral of 1/(sqrt(t^2-6t+13)) ?
The integral of 1/(sqrt(t^2-6t+13)) is ln(1/2 |t-3+sqrt(t^2-6t+13)|)+C