What is the integral of 1/(sqrt(t^2-2t+17)) ?

The integral of 1/(sqrt(t^2-2t+17)) is ln(1/4 |t-1+sqrt(t^2-2t+17)|)+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^2-2t+17))

\int \frac{1}{t ^{2} - 2 t + 17} d t

ln (\frac{1}{4} t - 1 + t^{2} - 2 t + 17) + C

Solution steps

\int \frac{1}{t ^{2} - 2 t + 17} d t

Complete the square

t^{2} - 2 t + 17 : (t - 1)^{2} + 16

= \int \frac{1}{( t - 1 ) ^{2} + 16} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} 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}{4} (t - 1))) + sec (arctan (\frac{1}{4} (t - 1)))

Simplify

ln tan (arctan (\frac{1}{4} (t - 1))) + sec (arctan (\frac{1}{4} (t - 1))) : ln (\frac{1}{4} t - 1 + t^{2} - 2 t + 17)

= ln (\frac{1}{4} t - 1 + t^{2} - 2 t + 17)

Add a constant to the solution

= ln (\frac{1}{4} t - 1 + t^{2} - 2 t + 17) + C

\frac{\partial}{\partial x} (7 x y - 3 x^{2} y)

x^{4} - 20 x^{2} + 64, at x = 2

2 t y^{^{'}} + 4 y = 3

t (t^{2} - 1)^{2}

\int 1 - r^{2} \cdot r d r

What is the integral of 1/(sqrt(t^2-2t+17)) ?
The integral of 1/(sqrt(t^2-2t+17)) is ln(1/4 |t-1+sqrt(t^2-2t+17)|)+C