What is the integral of x/(2x^2+8x+11) ?

The integral of x/(2x^2+8x+11) is 1/4 ln|2x^2+8x+11|-sqrt(2/3)arctan(sqrt(2/3)(x+2))+C

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integral of x/(2x^2+8x+11)

\int \frac{x}{2 x ^{2} + 8 x + 11} d x

\frac{1}{4} ln 2 x^{2} + 8 x + 11 - \frac{2}{3} arctan (\frac{2}{3} (x + 2)) + C

Solution steps

\int \frac{x}{2 x ^{2} + 8 x + 11} d x

Complete the square

2 x^{2} + 8 x + 11 : 2 (x + 2)^{2} + 3

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

Apply u-substitution

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

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

Expand

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

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

\int \frac{u}{2 u ^{2} + 3} d u = \frac{1}{4} ln 2 u^{2} + 3

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

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

Substitute back

u = x + 2

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

Simplify

\frac{1}{4} ln 2 (x + 2)^{2} + 3 - \frac{2}{3} arctan (\frac{2}{3} (x + 2)) : \frac{1}{4} ln 2 x^{2} + 8 x + 11 - \frac{2}{3} arctan (\frac{2}{3} (x + 2))

= \frac{1}{4} ln 2 x^{2} + 8 x + 11 - \frac{2}{3} arctan (\frac{2}{3} (x + 2))

Add a constant to the solution

= \frac{1}{4} ln 2 x^{2} + 8 x + 11 - \frac{2}{3} arctan (\frac{2}{3} (x + 2)) + C

\frac{\partial}{\partial z} (x^{y} + ln (z))

\int (2 x - 2) d x

\frac{\partial}{\partial x} (3 x^{2} y + y^{2} + yz)

\frac{d}{d x} (\frac{6 x}{x ^{2} - 3})

x \to - 5 lim (\frac{x ^{3} + 125}{x ^{2} - 25})

What is the integral of x/(2x^2+8x+11) ?
The integral of x/(2x^2+8x+11) is 1/4 ln|2x^2+8x+11|-sqrt(2/3)arctan(sqrt(2/3)(x+2))+C