What is the integral of (2x+1)/(x^2+2x+3) ?

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

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integral of (2x+1)/(x^2+2x+3)

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Expand

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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