integral (3x)/((x^2+4x+6))

Solución

integral

\frac{3 x}{( x ^{2} + 4 x + 6 )}

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

Pasos de solución

\int \frac{3 x}{x ^{2} + 4 x + 6} d x

Sacar la constante:

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

= 3 \cdot \int \frac{x}{x ^{2} + 4 x + 6} d x

Completar el cuadrado

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

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

Aplicar integración por sustitución

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

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

Aplicar integración por sustitución

= 3 \cdot \int \frac{v - 2}{v ^{2} + 1} d v

Expandir

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

Aplicar la regla de la suma:

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

= 3 (\int \frac{v}{v ^{2} + 1} d v - \int \frac{2}{v ^{2} + 1} d v)

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

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

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

Sustitución hacia atrás

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

Simplificar

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

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

Agregar una constante a la solución

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

\frac{12}{x ^{2}}

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

\frac{1}{( 4 + x )}

f (x) = \frac{1}{x ^{\frac{1}{2}}}

\frac{x + 1}{3 x + 2}