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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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