What is the integral of (x+17)/(x^2+16x+68) ?

The integral of (x+17)/(x^2+16x+68) is 1/2 ln|x^2+16x+68|+9/2 arctan((x+8)/2)+C

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integral of (x+17)/(x^2+16x+68)

\int \frac{x + 17}{x ^{2} + 16 x + 68} d x

\frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2}) + C

Solution steps

\int \frac{x + 17}{x ^{2} + 16 x + 68} d x

Apply u-substitution

: \int \frac{u}{u ^{2} - 18 u + 85} d u

= \int \frac{u}{u ^{2} - 18 u + 85} d u

Complete the square

u^{2} - 18 u + 85 : (u - 9)^{2} + 4

= \int \frac{u}{( u - 9 ) ^{2} + 4} d u

Apply u-substitution

: \int \frac{v + 9}{v ^{2} + 4} d v

= \int \frac{v + 9}{v ^{2} + 4} d v

Expand

\frac{v + 9}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{9}{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{9}{v ^{2} + 4} d v

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

\int \frac{9}{v ^{2} + 4} d v = \frac{9}{2} arctan (\frac{v}{2})

= \frac{1}{2} ln v^{2} + 4 + \frac{9}{2} arctan (\frac{v}{2})

Substitute back

= \frac{1}{2} ln (x + 17 - 9)^{2} + 4 + \frac{9}{2} arctan (\frac{x + 17 - 9}{2})

Simplify

\frac{1}{2} ln (x + 17 - 9)^{2} + 4 + \frac{9}{2} arctan (\frac{x + 17 - 9}{2}) : \frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2})

= \frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2})

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 16 x + 68 + \frac{9}{2} arctan (\frac{x + 8}{2}) + C

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What is the integral of (x+17)/(x^2+16x+68) ?
The integral of (x+17)/(x^2+16x+68) is 1/2 ln|x^2+16x+68|+9/2 arctan((x+8)/2)+C