What is the integral of (x+3)/(2x^2+8x+20) ?

The integral of (x+3)/(2x^2+8x+20) is 1/4 ln|2x^2+8x+20|+1/(2sqrt(6))arctan(1/(sqrt(6))(x+2))+C

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integral of (x+3)/(2x^2+8x+20)

\int \frac{x + 3}{2 x ^{2} + 8 x + 20} d x

\frac{1}{4} ln 2 x^{2} + 8 x + 20 + \frac{1}{2 6} arctan (\frac{1}{6} (x + 2)) + C

Solution steps

\int \frac{x + 3}{2 x ^{2} + 8 x + 20} d x

Complete the square

2 x^{2} + 8 x + 20 : 2 (x + 2)^{2} + 12

= \int \frac{x + 3}{2 ( x + 2 ) ^{2} + 12} d x

Apply u-substitution

: \int \frac{u + 1}{2 u ^{2} + 12} d u

= \int \frac{u + 1}{2 u ^{2} + 12} d u

Expand

\frac{u + 1}{2 u ^{2} + 12} : \frac{u}{2 u ^{2} + 12} + \frac{1}{2 u ^{2} + 12}

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}{2 u ^{2} + 12} d u + \int \frac{1}{2 u ^{2} + 12} d u

\int \frac{u}{2 u ^{2} + 12} d u = \frac{1}{4} ln 2 u^{2} + 12

\int \frac{1}{2 u ^{2} + 12} d u = \frac{1}{2 6} arctan (\frac{u}{6})

= \frac{1}{4} ln 2 u^{2} + 12 + \frac{1}{2 6} arctan (\frac{u}{6})

Substitute back

u = x + 2

= \frac{1}{4} ln 2 (x + 2)^{2} + 12 + \frac{1}{2 6} arctan (\frac{x + 2}{6})

Simplify

\frac{1}{4} ln 2 (x + 2)^{2} + 12 + \frac{1}{2 6} arctan (\frac{x + 2}{6}) : \frac{1}{4} ln 2 x^{2} + 8 x + 20 + \frac{1}{2 6} arctan (\frac{1}{6} (x + 2))

= \frac{1}{4} ln 2 x^{2} + 8 x + 20 + \frac{1}{2 6} arctan (\frac{1}{6} (x + 2))

Add a constant to the solution

= \frac{1}{4} ln 2 x^{2} + 8 x + 20 + \frac{1}{2 6} arctan (\frac{1}{6} (x + 2)) + C

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What is the integral of (x+3)/(2x^2+8x+20) ?
The integral of (x+3)/(2x^2+8x+20) is 1/4 ln|2x^2+8x+20|+1/(2sqrt(6))arctan(1/(sqrt(6))(x+2))+C