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

The integral of (x+1)/(x^2-4x+8) is 1/2 ln|x^2-4x+8|+3/2 arctan(1/2 (x-2))+C

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integral of (x+1)/(x^2-4x+8)

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

\frac{1}{2} ln x^{2} - 4 x + 8 + \frac{3}{2} arctan (\frac{1}{2} (x - 2)) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{u + 3}{u ^{2} + 4} d u

= \int \frac{u + 3}{u ^{2} + 4} d u

Expand

\frac{u + 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} + \frac{3}{u ^{2} + 4}

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} + 4} d u + \int \frac{3}{u ^{2} + 4} d u

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

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

= \frac{1}{2} ln u^{2} + 4 + \frac{3}{2} arctan (\frac{u}{2})

Substitute back

u = x - 2

= \frac{1}{2} ln (x - 2)^{2} + 4 + \frac{3}{2} arctan (\frac{x - 2}{2})

Simplify

\frac{1}{2} ln (x - 2)^{2} + 4 + \frac{3}{2} arctan (\frac{x - 2}{2}) : \frac{1}{2} ln x^{2} - 4 x + 8 + \frac{3}{2} arctan (\frac{1}{2} (x - 2))

= \frac{1}{2} ln x^{2} - 4 x + 8 + \frac{3}{2} arctan (\frac{1}{2} (x - 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 4 x + 8 + \frac{3}{2} arctan (\frac{1}{2} (x - 2)) + C

\int_{- 1}^{2} (2 - y^{2}) - (- y) d y

n = 0 \sum \infty 10 (\frac{1}{3})^{n}

(\frac{x + 5}{x ^{2} + 2})^{2}

(c_{2} sin (8 t))^{^{'}}

What is the integral of (x+1)/(x^2-4x+8) ?
The integral of (x+1)/(x^2-4x+8) is 1/2 ln|x^2-4x+8|+3/2 arctan(1/2 (x-2))+C