What is the integral of (x+9)/(x^2+8x+20) ?

The integral of (x+9)/(x^2+8x+20) is 1/2 ln|x^2+8x+20|+5/2 arctan((x+4)/2)+C

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integral of (x+9)/(x^2+8x+20)

\int \frac{x + 9}{x ^{2} + 8 x + 20} d x

\frac{1}{2} ln x^{2} + 8 x + 20 + \frac{5}{2} arctan (\frac{x + 4}{2}) + C

Solution steps

\int \frac{x + 9}{x ^{2} + 8 x + 20} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 10 u + 29} d u

= \int \frac{u}{u ^{2} - 10 u + 29} d u

Complete the square

u^{2} - 10 u + 29 : (u - 5)^{2} + 4

= \int \frac{u}{( u - 5 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v + 5}{v ^{2} + 4} d v

= \int \frac{v + 5}{v ^{2} + 4} d v

Expand

\frac{v + 5}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{5}{v ^{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{v}{v ^{2} + 4} d v + \int \frac{5}{v ^{2} + 4} d v

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

\int \frac{5}{v ^{2} + 4} d v = \frac{5}{2} arctan (\frac{v}{2})

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

Substitute back

= \frac{1}{2} ln (x + 9 - 5)^{2} + 4 + \frac{5}{2} arctan (\frac{x + 9 - 5}{2})

Simplify

\frac{1}{2} ln (x + 9 - 5)^{2} + 4 + \frac{5}{2} arctan (\frac{x + 9 - 5}{2}) : \frac{1}{2} ln x^{2} + 8 x + 20 + \frac{5}{2} arctan (\frac{x + 4}{2})

= \frac{1}{2} ln x^{2} + 8 x + 20 + \frac{5}{2} arctan (\frac{x + 4}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 8 x + 20 + \frac{5}{2} arctan (\frac{x + 4}{2}) + C

\frac{d}{d x} ((2 x + 5)^{\frac{1}{2}})

\frac{d}{d t} (6 cos (2 t))

y = x^{3}, y = x^{2} - 9 x, [- 1, 1]

\int (3 e^{5 x} - \frac{7}{x} + \frac{4}{1 + x ^{2}}) d x

What is the integral of (x+9)/(x^2+8x+20) ?
The integral of (x+9)/(x^2+8x+20) is 1/2 ln|x^2+8x+20|+5/2 arctan((x+4)/2)+C