What is the integral of 1/(sqrt(t^2-2t+37)) ?

The integral of 1/(sqrt(t^2-2t+37)) is ln(1/6 |t-1+sqrt(t^2-2t+37)|)+C

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integral of 1/(sqrt(t^2-2t+37))

\int \frac{1}{t ^{2} - 2 t + 37} d t

ln (\frac{1}{6} t - 1 + t^{2} - 2 t + 37) + C

Solution steps

\int \frac{1}{t ^{2} - 2 t + 37} d t

Complete the square

t^{2} - 2 t + 37 : (t - 1)^{2} + 36

= \int \frac{1}{( t - 1 ) ^{2} + 36} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 36} d u

= \int \frac{1}{u ^{2} + 36} 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}{6} (t - 1))) + sec (arctan (\frac{1}{6} (t - 1)))

Simplify

ln tan (arctan (\frac{1}{6} (t - 1))) + sec (arctan (\frac{1}{6} (t - 1))) : ln (\frac{1}{6} t - 1 + t^{2} - 2 t + 37)

= ln (\frac{1}{6} t - 1 + t^{2} - 2 t + 37)

Add a constant to the solution

= ln (\frac{1}{6} t - 1 + t^{2} - 2 t + 37) + C

\int_{0}^{\frac{π}{4}} 7 sec^{2} (x) d x

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

y^{^{'}} = \frac{y ^{2}}{t ^{3}}

ln (x e^{- 6 x})

\frac{sec ( θ )}{1 + sec ( θ )}

What is the integral of 1/(sqrt(t^2-2t+37)) ?
The integral of 1/(sqrt(t^2-2t+37)) is ln(1/6 |t-1+sqrt(t^2-2t+37)|)+C