What is the integral of x/(x^2+6x+13) ?

The integral of x/(x^2+6x+13) is 1/2 ln|x^2+6x+13|-3/2 arctan(1/2 (x+3))+C

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integral of x/(x^2+6x+13)

\int \frac{x}{x ^{2} + 6 x + 13} d x

\frac{1}{2} ln x^{2} + 6 x + 13 - \frac{3}{2} arctan (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{x}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= \int \frac{x}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{u - 3}{u ^{2} + 4} d u

= \int \frac{u - 3}{u ^{2} + 4} d u

Expand

\frac{u - 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{3}{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{3}{u ^{2} + 4} d u

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

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

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

Substitute back

u = x + 3

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

Simplify

\frac{1}{2} ln (x + 3)^{2} + 4 - \frac{3}{2} arctan (\frac{x + 3}{2}) : \frac{1}{2} ln x^{2} + 6 x + 13 - \frac{3}{2} arctan (\frac{1}{2} (x + 3))

= \frac{1}{2} ln x^{2} + 6 x + 13 - \frac{3}{2} arctan (\frac{1}{2} (x + 3))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 6 x + 13 - \frac{3}{2} arctan (\frac{1}{2} (x + 3)) + C

\int x e^{- x} sin (x) d x

x \to - \infty lim (x^{4} e^{x})

\frac{d}{d x} (1 - 2 e^{3 x})

f (x) = (\frac{3 x ^{2}}{2 x + 4})^{3}, x = 2

100 x + 0.01 x^{2}

What is the integral of x/(x^2+6x+13) ?
The integral of x/(x^2+6x+13) is 1/2 ln|x^2+6x+13|-3/2 arctan(1/2 (x+3))+C