What is the integral of 1/(sqrt(t^2-4t+13)) ?

The integral of 1/(sqrt(t^2-4t+13)) is ln(1/3 |t-2+sqrt(t^2-4t+13)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(t^2-4t+13))

\int \frac{1}{t ^{2} - 4 t + 13} d t

ln (\frac{1}{3} t - 2 + t^{2} - 4 t + 13) + C

Solution steps

\int \frac{1}{t ^{2} - 4 t + 13} d t

Complete the square

t^{2} - 4 t + 13 : (t - 2)^{2} + 9

= \int \frac{1}{( t - 2 ) ^{2} + 9} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} 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}{3} (t - 2))) + sec (arctan (\frac{1}{3} (t - 2)))

Simplify

ln tan (arctan (\frac{1}{3} (t - 2))) + sec (arctan (\frac{1}{3} (t - 2))) : ln (\frac{1}{3} t - 2 + t^{2} - 4 t + 13)

= ln (\frac{1}{3} t - 2 + t^{2} - 4 t + 13)

Add a constant to the solution

= ln (\frac{1}{3} t - 2 + t^{2} - 4 t + 13) + C

y = ln (1 - x^{2} + 2 x^{4}), at x = 1

x \to 5 lim (6 - x - x^{2})

\int 400 - 0.4 x d x

\frac{d}{d y} (x^{2} + y^{2})

n = 1 \sum \infty n^{2} (- 1)^{n}

What is the integral of 1/(sqrt(t^2-4t+13)) ?
The integral of 1/(sqrt(t^2-4t+13)) is ln(1/3 |t-2+sqrt(t^2-4t+13)|)+C