What is the integral of 1/(sqrt(t^2-4t+20)) ?

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

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integral of 1/(sqrt(t^2-4t+20))

\int \frac{1}{t ^{2} - 4 t + 20} d t

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

Solution steps

\int \frac{1}{t ^{2} - 4 t + 20} d t

Complete the square

t^{2} - 4 t + 20 : (t - 2)^{2} + 16

= \int \frac{1}{( t - 2 ) ^{2} + 16} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} 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}{4} (t - 2))) + sec (arctan (\frac{1}{4} (t - 2)))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{ln ( e ^{x} )}{x})

\int 6 x^{2} + 3 d x

\frac{\partial}{\partial x} (6 x + y)

\int x^{2} + 1 d x

\frac{d}{d x} (\frac{6 x + 4 x ^{2}}{( 3 + 4 x ) ^{2}})

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