What is the integral of 1/(sqrt(t^2-12t+45)) ?

The integral of 1/(sqrt(t^2-12t+45)) is ln(1/3 |t-6+sqrt(t^2-12t+45)|)+C

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integral of 1/(sqrt(t^2-12t+45))

\int \frac{1}{t ^{2} - 12 t + 45} d t

ln (\frac{1}{3} t - 6 + t^{2} - 12 t + 45) + C

Solution steps

\int \frac{1}{t ^{2} - 12 t + 45} d t

Complete the square

t^{2} - 12 t + 45 : (t - 6)^{2} + 9

= \int \frac{1}{( t - 6 ) ^{2} + 9} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \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 - 6))) + sec (arctan (\frac{1}{3} (t - 6)))

Simplify

ln tan (arctan (\frac{1}{3} (t - 6))) + sec (arctan (\frac{1}{3} (t - 6))) : ln (\frac{1}{3} t - 6 + t^{2} - 12 t + 45)

= ln (\frac{1}{3} t - 6 + t^{2} - 12 t + 45)

Add a constant to the solution

= ln (\frac{1}{3} t - 6 + t^{2} - 12 t + 45) + C

(1 + x) e^{- x}

\frac{e ^{x} - e ^{- x}}{2}

\frac{d}{d x} ((x)^{\frac{1}{3}})

\frac{d}{d x} (- \frac{ln ( 1 + 6 x )}{x})

\frac{d}{d x} (\frac{1}{x ( x - 1 )})

What is the integral of 1/(sqrt(t^2-12t+45)) ?
The integral of 1/(sqrt(t^2-12t+45)) is ln(1/3 |t-6+sqrt(t^2-12t+45)|)+C