What is the integral of 1/(x^2-16x+68) ?

The integral of 1/(x^2-16x+68) is 1/2 arctan((x-8)/2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x^2-16x+68)

\int \frac{1}{x ^{2} - 16 x + 68} d x

\frac{1}{2} arctan (\frac{x - 8}{2}) + C

Solution steps

\int \frac{1}{x ^{2} - 16 x + 68} d x

Complete the square

x^{2} - 16 x + 68 : (x - 8)^{2} + 4

= \int \frac{1}{( x - 8 ) ^{2} + 4} d x

Apply u-substitution

= \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{x - 8}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x - 8}{2}) + C

What is the integral of 1/(x^2-16x+68) ?
The integral of 1/(x^2-16x+68) is 1/2 arctan((x-8)/2)+C