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

The integral of x/(x^2-2x+8) is 1/2 ln|x^2-2x+8|+1/(sqrt(7))arctan(1/(sqrt(7))(x-1))+C

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

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

\frac{1}{2} ln x^{2} - 2 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (x - 1)) + C

Solution steps

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

Complete the square

x^{2} - 2 x + 8 : (x - 1)^{2} + 7

= \int \frac{x}{( x - 1 ) ^{2} + 7} d x

Apply u-substitution

: \int \frac{u + 1}{u ^{2} + 7} d u

= \int \frac{u + 1}{u ^{2} + 7} d u

Expand

\frac{u + 1}{u ^{2} + 7} : \frac{u}{u ^{2} + 7} + \frac{1}{u ^{2} + 7}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{u ^{2} + 7} d u + \int \frac{1}{u ^{2} + 7} d u

\int \frac{u}{u ^{2} + 7} d u = \frac{1}{2} ln u^{2} + 7

\int \frac{1}{u ^{2} + 7} d u = \frac{1}{7} arctan (\frac{u}{7})

= \frac{1}{2} ln u^{2} + 7 + \frac{1}{7} arctan (\frac{u}{7})

Substitute back

u = x - 1

= \frac{1}{2} ln (x - 1)^{2} + 7 + \frac{1}{7} arctan (\frac{x - 1}{7})

Simplify

\frac{1}{2} ln (x - 1)^{2} + 7 + \frac{1}{7} arctan (\frac{x - 1}{7}) : \frac{1}{2} ln x^{2} - 2 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (x - 1))

= \frac{1}{2} ln x^{2} - 2 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (x - 1))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 2 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (x - 1)) + C

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What is the integral of x/(x^2-2x+8) ?
The integral of x/(x^2-2x+8) is 1/2 ln|x^2-2x+8|+1/(sqrt(7))arctan(1/(sqrt(7))(x-1))+C