What is the integral of 1/(sqrt(t^2-4t+29)) ?

The integral of 1/(sqrt(t^2-4t+29)) is ln(1/5 |t-2+sqrt(t^2-4t+29)|)+C

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integral of 1/(sqrt(t^2-4t+29))

\int \frac{1}{t ^{2} - 4 t + 29} d t

ln (\frac{1}{5} t - 2 + t^{2} - 4 t + 29) + C

Solution steps

\int \frac{1}{t ^{2} - 4 t + 29} d t

Complete the square

t^{2} - 4 t + 29 : (t - 2)^{2} + 25

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

Simplify

ln tan (arctan (\frac{1}{5} (t - 2))) + sec (arctan (\frac{1}{5} (t - 2))) : ln (\frac{1}{5} t - 2 + t^{2} - 4 t + 29)

= ln (\frac{1}{5} t - 2 + t^{2} - 4 t + 29)

Add a constant to the solution

= ln (\frac{1}{5} t - 2 + t^{2} - 4 t + 29) + C

\int 4 - t d t

\int_{0}^{2} (x^{2} - 2)^{4} x d x

y = 4 x, y = \frac{5}{4} x, y = 21 - x^{2}

x \to - 1 - lim (\frac{x ^{3}}{3 x + 3})

3 x^{4}

What is the integral of 1/(sqrt(t^2-4t+29)) ?
The integral of 1/(sqrt(t^2-4t+29)) is ln(1/5 |t-2+sqrt(t^2-4t+29)|)+C