What is the integral of x/(2x^2+12x+23) ?

The integral of x/(2x^2+12x+23) is 1/4 ln|2x^2+12x+23|-3/(sqrt(10))arctan(sqrt(2/5)(x+3))+C

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integral of x/(2x^2+12x+23)

\int \frac{x}{2 x ^{2} + 12 x + 23} d x

\frac{1}{4} ln 2 x^{2} + 12 x + 23 - \frac{3}{10} arctan (\frac{2}{5} (x + 3)) + C

Solution steps

\int \frac{x}{2 x ^{2} + 12 x + 23} d x

Complete the square

2 x^{2} + 12 x + 23 : 2 (x + 3)^{2} + 5

= \int \frac{x}{2 ( x + 3 ) ^{2} + 5} d x

Apply u-substitution

: \int \frac{u - 3}{2 u ^{2} + 5} d u

= \int \frac{u - 3}{2 u ^{2} + 5} d u

Expand

\frac{u - 3}{2 u ^{2} + 5} : \frac{u}{2 u ^{2} + 5} - \frac{3}{2 u ^{2} + 5}

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} + 5} d u - \int \frac{3}{2 u ^{2} + 5} d u

\int \frac{u}{2 u ^{2} + 5} d u = \frac{1}{4} ln 2 u^{2} + 5

\int \frac{3}{2 u ^{2} + 5} d u = \frac{3}{10} arctan (\frac{2}{5} u)

= \frac{1}{4} ln 2 u^{2} + 5 - \frac{3}{10} arctan (\frac{2}{5} u)

Substitute back

u = x + 3

= \frac{1}{4} ln 2 (x + 3)^{2} + 5 - \frac{3}{10} arctan (\frac{2}{5} (x + 3))

Simplify

\frac{1}{4} ln 2 (x + 3)^{2} + 5 - \frac{3}{10} arctan (\frac{2}{5} (x + 3)) : \frac{1}{4} ln 2 x^{2} + 12 x + 23 - \frac{3}{10} arctan (\frac{2}{5} (x + 3))

= \frac{1}{4} ln 2 x^{2} + 12 x + 23 - \frac{3}{10} arctan (\frac{2}{5} (x + 3))

Add a constant to the solution

= \frac{1}{4} ln 2 x^{2} + 12 x + 23 - \frac{3}{10} arctan (\frac{2}{5} (x + 3)) + C

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What is the integral of x/(2x^2+12x+23) ?
The integral of x/(2x^2+12x+23) is 1/4 ln|2x^2+12x+23|-3/(sqrt(10))arctan(sqrt(2/5)(x+3))+C