What is the integral of 1/(sqrt(t^2-6t+34)) ?

The integral of 1/(sqrt(t^2-6t+34)) is ln(1/5 |t-3+sqrt(t^2-6t+34)|)+C

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integral of 1/(sqrt(t^2-6t+34))

\int \frac{1}{t ^{2} - 6 t + 34} d t

ln (\frac{1}{5} t - 3 + t^{2} - 6 t + 34) + C

Solution steps

\int \frac{1}{t ^{2} - 6 t + 34} d t

Complete the square

t^{2} - 6 t + 34 : (t - 3)^{2} + 25

= \int \frac{1}{( t - 3 ) ^{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 - 3))) + sec (arctan (\frac{1}{5} (t - 3)))

Simplify

ln tan (arctan (\frac{1}{5} (t - 3))) + sec (arctan (\frac{1}{5} (t - 3))) : ln (\frac{1}{5} t - 3 + t^{2} - 6 t + 34)

= ln (\frac{1}{5} t - 3 + t^{2} - 6 t + 34)

Add a constant to the solution

= ln (\frac{1}{5} t - 3 + t^{2} - 6 t + 34) + C

y = \frac{1}{x}

f (x) = 8 (x - \frac{1}{x})^{2}

\int \frac{3}{4} d x

\int_{2}^{4} 2 π x (5 - x) d x

\int sin (x^{3}) d x

What is the integral of 1/(sqrt(t^2-6t+34)) ?
The integral of 1/(sqrt(t^2-6t+34)) is ln(1/5 |t-3+sqrt(t^2-6t+34)|)+C