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

The integral of x/(x^2+2x+17) is 1/2 ln|x^2+2x+17|-1/4 arctan(1/4 (x+1))+C

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

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

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

Solution steps

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

Complete the square

x^{2} + 2 x + 17 : (x + 1)^{2} + 16

= \int \frac{x}{( x + 1 ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{u - 1}{u ^{2} + 16} d u

Expand

\frac{u - 1}{u ^{2} + 16} : \frac{u}{u ^{2} + 16} - \frac{1}{u ^{2} + 16}

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}{u ^{2} + 16} d u - \int \frac{1}{u ^{2} + 16} d u

\int \frac{u}{u ^{2} + 16} d u = \frac{1}{2} ln u^{2} + 16

\int \frac{1}{u ^{2} + 16} d u = \frac{1}{4} arctan (\frac{u}{4})

= \frac{1}{2} ln u^{2} + 16 - \frac{1}{4} arctan (\frac{u}{4})

Substitute back

u = x + 1

= \frac{1}{2} ln (x + 1)^{2} + 16 - \frac{1}{4} arctan (\frac{x + 1}{4})

Simplify

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

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

Add a constant to the solution

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

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What is the integral of x/(x^2+2x+17) ?
The integral of x/(x^2+2x+17) is 1/2 ln|x^2+2x+17|-1/4 arctan(1/4 (x+1))+C