What is the integral of 1/(sqrt(t^2-8t+41)) ?

The integral of 1/(sqrt(t^2-8t+41)) is ln(1/5 |t-4+sqrt(t^2-8t+41)|)+C

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integral of 1/(sqrt(t^2-8t+41))

\int \frac{1}{t ^{2} - 8 t + 41} d t

ln (\frac{1}{5} t - 4 + t^{2} - 8 t + 41) + C

Solution steps

\int \frac{1}{t ^{2} - 8 t + 41} d t

Complete the square

t^{2} - 8 t + 41 : (t - 4)^{2} + 25

= \int \frac{1}{( t - 4 ) ^{2} + 25} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 25} d u

= \int \frac{1}{u ^{2} + 25} 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}{5} (t - 4))) + sec (arctan (\frac{1}{5} (t - 4)))

Simplify

ln tan (arctan (\frac{1}{5} (t - 4))) + sec (arctan (\frac{1}{5} (t - 4))) : ln (\frac{1}{5} t - 4 + t^{2} - 8 t + 41)

= ln (\frac{1}{5} t - 4 + t^{2} - 8 t + 41)

Add a constant to the solution

= ln (\frac{1}{5} t - 4 + t^{2} - 8 t + 41) + C

\frac{d}{d x} (8 x (x^{5} + 1)^{3})

\frac{\partial}{\partial x} (4 + x + x^{2} y)

x \to 0 lim (\frac{4}{x})

\frac{d}{d x} (2 x^{2} - 6)

e^{- 3 s}

What is the integral of 1/(sqrt(t^2-8t+41)) ?
The integral of 1/(sqrt(t^2-8t+41)) is ln(1/5 |t-4+sqrt(t^2-8t+41)|)+C