What is the integral of (2x-3)/(x^2+6x+13) ?

The integral of (2x-3)/(x^2+6x+13) is ln| 1/4 (x+3)^2+1|-9/2 arctan(1/2 (x+3))+C

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integral of (2x-3)/(x^2+6x+13)

\int \frac{2 x - 3}{x ^{2} + 6 x + 13} d x

ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} + 6 x + 13} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 9}{u ^{2} + 4} d u

= \int \frac{2 u - 9}{u ^{2} + 4} d u

Apply Integral Substitution

= \int \frac{4 v - 9}{2 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{4 v - 9}{v ^{2} + 1} d v

Expand

\frac{4 v - 9}{v ^{2} + 1} : \frac{4 v}{v ^{2} + 1} - \frac{9}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{4 v}{v ^{2} + 1} d v - \int \frac{9}{v ^{2} + 1} d v)

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

\int \frac{9}{v ^{2} + 1} d v = 9 arctan (v)

= \frac{1}{2} (2 ln v^{2} + 1 - 9 arctan (v))

Substitute back

= \frac{1}{2} (2 ln (\frac{x + 3}{2})^{2} + 1 - 9 arctan (\frac{x + 3}{2}))

Simplify

\frac{1}{2} (2 ln (\frac{x + 3}{2})^{2} + 1 - 9 arctan (\frac{x + 3}{2})) : ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3))

= ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3))

Add a constant to the solution

= ln \frac{1}{4} (x + 3)^{2} + 1 - \frac{9}{2} arctan (\frac{1}{2} (x + 3)) + C

\frac{x ^{2}}{3}

(- 12, 5), (- 4, - 1)

f (x) = x^{2} - 12, g (x) = x - 6

\frac{\partial}{\partial x} (sin (4 x^{3} y - 6 x y^{2}))

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

What is the integral of (2x-3)/(x^2+6x+13) ?
The integral of (2x-3)/(x^2+6x+13) is ln| 1/4 (x+3)^2+1|-9/2 arctan(1/2 (x+3))+C