What is the integral of 1/(sqrt(t^2-2t+5)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(sqrt(t^2-2t+5))

\int \frac{1}{t ^{2} - 2 t + 5} d t

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

Solution steps

\int \frac{1}{t ^{2} - 2 t + 5} d t

Complete the square

t^{2} - 2 t + 5 : (t - 1)^{2} + 4

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

Apply u-substitution

= \int \frac{1}{u ^{2} + 4} d u

Apply Trigonometric Substitution

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

Simplify

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

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

Add a constant to the solution

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

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