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

The integral of 1/(s^2-2s+5) is 1/2 arctan((s-1)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(s^2-2s+5)

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

\frac{1}{2} arctan (\frac{s - 1}{2}) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} arctan (\frac{s - 1}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{s - 1}{2}) + C

x \to - 4 lim (\frac{4 - ∣ x ∣}{4 + x})

\frac{d}{d x} (csec x)

\frac{\partial}{\partial x} (5 - x^{2} + 3 y^{2})

3 x^{2}, at x = 2

x \to 0 lim (\frac{1 + x}{x})

What is the integral of 1/(s^2-2s+5) ?
The integral of 1/(s^2-2s+5) is 1/2 arctan((s-1)/2)+C