What is the integral of 1/(sqrt(t^2-8t+25)) ?

The integral of 1/(sqrt(t^2-8t+25)) is ln(1/3 |t-4+sqrt(t^2-8t+25)|)+C

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integral of 1/(sqrt(t^2-8t+25))

\int \frac{1}{t ^{2} - 8 t + 25} d t

ln (\frac{1}{3} t - 4 + t^{2} - 8 t + 25) + C

Solution steps

\int \frac{1}{t ^{2} - 8 t + 25} d t

Complete the square

t^{2} - 8 t + 25 : (t - 4)^{2} + 9

= \int \frac{1}{( t - 4 ) ^{2} + 9} d t

Apply u-substitution

= \int \frac{1}{u ^{2} + 9} 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}{3} (t - 4))) + sec (arctan (\frac{1}{3} (t - 4)))

Simplify

ln tan (arctan (\frac{1}{3} (t - 4))) + sec (arctan (\frac{1}{3} (t - 4))) : ln (\frac{1}{3} t - 4 + t^{2} - 8 t + 25)

= ln (\frac{1}{3} t - 4 + t^{2} - 8 t + 25)

Add a constant to the solution

= ln (\frac{1}{3} t - 4 + t^{2} - 8 t + 25) + C

What is the integral of 1/(sqrt(t^2-8t+25)) ?
The integral of 1/(sqrt(t^2-8t+25)) is ln(1/3 |t-4+sqrt(t^2-8t+25)|)+C