What is the integral of 1/(sqrt(t^2-4t+40)) ?

The integral of 1/(sqrt(t^2-4t+40)) is ln(1/6 |t-2+sqrt(t^2-4t+40)|)+C

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integral of 1/(sqrt(t^2-4t+40))

\int \frac{1}{t ^{2} - 4 t + 40} d t

ln (\frac{1}{6} t - 2 + t^{2} - 4 t + 40) + C

Solution steps

\int \frac{1}{t ^{2} - 4 t + 40} d t

Complete the square

t^{2} - 4 t + 40 : (t - 2)^{2} + 36

= \int \frac{1}{( t - 2 ) ^{2} + 36} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 36} d u

= \int \frac{1}{u ^{2} + 36} 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}{6} (t - 2))) + sec (arctan (\frac{1}{6} (t - 2)))

Simplify

ln tan (arctan (\frac{1}{6} (t - 2))) + sec (arctan (\frac{1}{6} (t - 2))) : ln (\frac{1}{6} t - 2 + t^{2} - 4 t + 40)

= ln (\frac{1}{6} t - 2 + t^{2} - 4 t + 40)

Add a constant to the solution

= ln (\frac{1}{6} t - 2 + t^{2} - 4 t + 40) + C

\int \frac{20 - 20 x}{1 - x} d x

\int ln (5 - x) d x

f (x) = 2 x^{2} - 7 x + 5

\int \frac{x ^{3} + 2 x ^{2} - 18 x + 2}{x ^{2} + x - 12} d x

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 7

What is the integral of 1/(sqrt(t^2-4t+40)) ?
The integral of 1/(sqrt(t^2-4t+40)) is ln(1/6 |t-2+sqrt(t^2-4t+40)|)+C