What is the integral of 1/(x^2-8x+20) ?

The integral of 1/(x^2-8x+20) is 1/2 arctan((x-4)/2)+C

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integral of 1/(x^2-8x+20)

\int \frac{1}{x ^{2} - 8 x + 20} d x

\frac{1}{2} arctan (\frac{x - 4}{2}) + C

Solution steps

\int \frac{1}{x ^{2} - 8 x + 20} d x

Complete the square

x^{2} - 8 x + 20 : (x - 4)^{2} + 4

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

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{x - 4}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x - 4}{2}) + C

\int e^{- s x} d x

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

\frac{d}{d x} (\frac{x ^{2} + x + 1}{2 x})

\frac{d y}{d x} = 2 x + y, y (0) = 4

\frac{d}{d x} (6 e^{- 8 x})

What is the integral of 1/(x^2-8x+20) ?
The integral of 1/(x^2-8x+20) is 1/2 arctan((x-4)/2)+C