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

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

y^{^{'''}} - y^{^{''}} - 20 y^{^{'}} = 0

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

2 x^{3} + 36 x^{2} + 210 x + 14

\frac{d}{d x} (e^{- x} (x sin (y) - y cos (y)))

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