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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Complete the square

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

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

Apply u-substitution

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

= 2 \cdot \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

= 2 (\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})

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

2 x^{^{''}} + 7 x^{^{'}} + 3 x = 0, x (0) = 3, x^{^{'}} (0) = 0

(1 + ln (x) + \frac{y}{x}) d x = (8 - ln (x)) d y

\frac{d}{d x} (ln (sinh (2 x)))

\int \frac{e ^{x}}{1 + 9 e ^{2 x}} d x

x \to 3 lim (2 e^{x - 3})

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