What is the integral of (3x+8)/(x^2+6x+14) ?

The integral of (3x+8)/(x^2+6x+14) is 3/2 ln|x^2+6x+14|-1/(sqrt(5))arctan(1/(sqrt(5))(x+3))+C

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integral of (3x+8)/(x^2+6x+14)

\int \frac{3 x + 8}{x ^{2} + 6 x + 14} d x

\frac{3}{2} ln x^{2} + 6 x + 14 - \frac{1}{5} arctan (\frac{1}{5} (x + 3)) + C

Solution steps

\int \frac{3 x + 8}{x ^{2} + 6 x + 14} d x

Complete the square

x^{2} + 6 x + 14 : (x + 3)^{2} + 5

= \int \frac{3 x + 8}{( x + 3 ) ^{2} + 5} d x

Apply u-substitution

= \int \frac{3 u - 1}{u ^{2} + 5} d u

Expand

\frac{3 u - 1}{u ^{2} + 5} : \frac{3 u}{u ^{2} + 5} - \frac{1}{u ^{2} + 5}

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{3 u}{u ^{2} + 5} d u - \int \frac{1}{u ^{2} + 5} d u

\int \frac{3 u}{u ^{2} + 5} d u = \frac{3}{2} ln u^{2} + 5

\int \frac{1}{u ^{2} + 5} d u = \frac{1}{5} arctan (\frac{u}{5})

= \frac{3}{2} ln u^{2} + 5 - \frac{1}{5} arctan (\frac{u}{5})

Substitute back

u = x + 3

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

Simplify

\frac{3}{2} ln (x + 3)^{2} + 5 - \frac{1}{5} arctan (\frac{x + 3}{5}) : \frac{3}{2} ln x^{2} + 6 x + 14 - \frac{1}{5} arctan (\frac{1}{5} (x + 3))

= \frac{3}{2} ln x^{2} + 6 x + 14 - \frac{1}{5} arctan (\frac{1}{5} (x + 3))

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 6 x + 14 - \frac{1}{5} arctan (\frac{1}{5} (x + 3)) + C

What is the integral of (3x+8)/(x^2+6x+14) ?
The integral of (3x+8)/(x^2+6x+14) is 3/2 ln|x^2+6x+14|-1/(sqrt(5))arctan(1/(sqrt(5))(x+3))+C