What is the integral of (x-3)/(x^2+4x+8) ?

The integral of (x-3)/(x^2+4x+8) is 1/2 ln| 1/4 x^2+x+2|-5/2 arctan(1/2 x+1)+C

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integral of (x-3)/(x^2+4x+8)

\int \frac{x - 3}{x ^{2} + 4 x + 8} d x

\frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1) + C

Solution steps

\int \frac{x - 3}{x ^{2} + 4 x + 8} d x

Expand

\frac{x - 3}{x ^{2} + 4 x + 8} : \frac{x}{x ^{2} + 4 x + 8} - \frac{3}{x ^{2} + 4 x + 8}

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{x}{x ^{2} + 4 x + 8} d x - \int \frac{3}{x ^{2} + 4 x + 8} d x

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

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

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

Simplify

\frac{1}{2} ln \frac{1}{4} (x + 2)^{2} + 1 - arctan (\frac{1}{2} (x + 2)) - \frac{3}{2} arctan (\frac{1}{2} (x + 2)) : \frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1)

= \frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1)

Add a constant to the solution

= \frac{1}{2} ln \frac{1}{4} x^{2} + x + 2 - \frac{5}{2} arctan (\frac{1}{2} x + 1) + C

\int_{0}^{1} e^{3 x} d x

y = 11 x^{5} - 2 x^{3} + 6 x - 8

4 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

2^{x}, 4^{x}, \frac{1}{x}

\int e^{x} cos (8 x) d x

What is the integral of (x-3)/(x^2+4x+8) ?
The integral of (x-3)/(x^2+4x+8) is 1/2 ln| 1/4 x^2+x+2|-5/2 arctan(1/2 x+1)+C