What is the integral of x/(x^2+2x+5) ?

The integral of x/(x^2+2x+5) is 1/2 ln|x^2+2x+5|-1/2 arctan(1/2 (x+1))+C

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integral of x/(x^2+2x+5)

\int \frac{x}{x ^{2} + 2 x + 5} d x

\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

Solution steps

\int \frac{x}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

= \int \frac{x}{( x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 4} d u

= \int \frac{u - 1}{u ^{2} + 4} d u

Expand

\frac{u - 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{1}{u ^{2} + 4}

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} + 4} d u - \int \frac{1}{u ^{2} + 4} d u

\int \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln u^{2} + 4

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

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

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} ln (x + 1)^{2} + 4 - \frac{1}{2} arctan (\frac{x + 1}{2}) : \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))

= \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)) + C

x \to 0 lim (\frac{e ^{3 x - 1}}{x})

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

\int_{0}^{2} x x^{2} + 1 d x

\frac{3 s}{( s + 3 ) ^{2} ( s + 6 )}

\int_{0}^{l n (5)} 4 (5 - x e^{x}) d x

What is the integral of x/(x^2+2x+5) ?
The integral of x/(x^2+2x+5) is 1/2 ln|x^2+2x+5|-1/2 arctan(1/2 (x+1))+C