What is the integral of x/(x^2+2x+4) ?

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

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integral of x/(x^2+2x+4)

\int \frac{x}{x ^{2} + 2 x + 4} d x

\frac{1}{2} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{x}{x ^{2} + 2 x + 4} d x

Complete the square

x^{2} + 2 x + 4 : (x + 1)^{2} + 3

= \int \frac{x}{( x + 1 ) ^{2} + 3} d x

Apply u-substitution

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

Expand

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

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

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

\int \frac{1}{u ^{2} + 3} d u = \frac{1}{3} arctan (\frac{u}{3})

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

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} ln (x + 1)^{2} + 3 - \frac{1}{3} arctan (\frac{x + 1}{3}) : \frac{1}{2} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

= \frac{1}{2} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

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