What is the integral of 1/(sqrt(t^2-8t+20)) ?

The integral of 1/(sqrt(t^2-8t+20)) is ln(1/2 |t-4+sqrt(t^2-8t+20)|)+C

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integral of 1/(sqrt(t^2-8t+20))

\int \frac{1}{t ^{2} - 8 t + 20} d t

ln (\frac{1}{2} t - 4 + t^{2} - 8 t + 20) + C

Solution steps

\int \frac{1}{t ^{2} - 8 t + 20} d t

Complete the square

t^{2} - 8 t + 20 : (t - 4)^{2} + 4

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

Simplify

ln tan (arctan (\frac{1}{2} (t - 4))) + sec (arctan (\frac{1}{2} (t - 4))) : ln (\frac{1}{2} t - 4 + t^{2} - 8 t + 20)

= ln (\frac{1}{2} t - 4 + t^{2} - 8 t + 20)

Add a constant to the solution

= ln (\frac{1}{2} t - 4 + t^{2} - 8 t + 20) + C

x \to 0 lim (\frac{1}{cos ( 3 x )})

y^{^{''}} + 7 y^{^{'}} - 8 y = 0

\frac{d y}{d x} = 6 x^{5} e^{- y}

\frac{d}{d x} (sin (1 + x^{2}))

y = \frac{2 x ^{2} + 6 x + 4}{x}

What is the integral of 1/(sqrt(t^2-8t+20)) ?
The integral of 1/(sqrt(t^2-8t+20)) is ln(1/2 |t-4+sqrt(t^2-8t+20)|)+C