What is the integral of ((x-2))/(x^2-4x+9) ?

The integral of ((x-2))/(x^2-4x+9) is 1/2 ln|x^2-4x+9|+C

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integral of ((x-2))/(x^2-4x+9)

\int \frac{( x - 2 )}{x ^{2} - 4 x + 9} d x

\frac{1}{2} ln x^{2} - 4 x + 9 + C

Solution steps

\int \frac{x - 2}{x ^{2} - 4 x + 9} d x

Complete the square

x^{2} - 4 x + 9 : (x - 2)^{2} + 5

= \int \frac{x - 2}{( x - 2 ) ^{2} + 5} d x

Apply u-substitution

: \int \frac{u}{u ^{2} + 5} d u

= \int \frac{u}{u ^{2} + 5} d u

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} 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} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln (x - 2)^{2} + 5

Simplify

\frac{1}{2} ln (x - 2)^{2} + 5 : \frac{1}{2} ln x^{2} - 4 x + 9

= \frac{1}{2} ln x^{2} - 4 x + 9

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 4 x + 9 + C

y^{^{''}} + 2 y^{^{'}} = - 4 sin (2 x)

\int \frac{1}{x ^{5} - x ^{2}} d x

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

x \to 0 lim (\frac{- x ^{3}}{x ^{3}})

\int \frac{x}{9 - 2 x ^{4}} d x

What is the integral of ((x-2))/(x^2-4x+9) ?
The integral of ((x-2))/(x^2-4x+9) is 1/2 ln|x^2-4x+9|+C