What is the integral of 1/(sqrt(t^2-12t+40)) ?

The integral of 1/(sqrt(t^2-12t+40)) is ln(1/2 |t-6+sqrt(t^2-12t+40)|)+C

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integral of 1/(sqrt(t^2-12t+40))

\int \frac{1}{t ^{2} - 12 t + 40} d t

ln (\frac{1}{2} t - 6 + t^{2} - 12 t + 40) + C

Solution steps

\int \frac{1}{t ^{2} - 12 t + 40} d t

Complete the square

t^{2} - 12 t + 40 : (t - 6)^{2} + 4

= \int \frac{1}{( t - 6 ) ^{2} + 4} d t

Apply u-substitution

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

= \int \frac{1}{u ^{2} + 4} 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}{2} (t - 6))) + sec (arctan (\frac{1}{2} (t - 6)))

Simplify

ln tan (arctan (\frac{1}{2} (t - 6))) + sec (arctan (\frac{1}{2} (t - 6))) : ln (\frac{1}{2} t - 6 + t^{2} - 12 t + 40)

= ln (\frac{1}{2} t - 6 + t^{2} - 12 t + 40)

Add a constant to the solution

= ln (\frac{1}{2} t - 6 + t^{2} - 12 t + 40) + C

x \to 0 - lim \frac{1}{\frac{1}{x}}

(t^{6} + 26 t^{4} - t^{3} + 179)^{73}

x \to 0 lim ((1 + \frac{4}{x})^{x})

(0, 32), (100, 212)

\frac{d}{d t} (\frac{e ^{t}}{4 - 3 e ^{t}})

What is the integral of 1/(sqrt(t^2-12t+40)) ?
The integral of 1/(sqrt(t^2-12t+40)) is ln(1/2 |t-6+sqrt(t^2-12t+40)|)+C