integral de (33x)/(x^2+8x+7)

Solución

\int \frac{33 x}{x ^{2} + 8 x + 7} d x

33 (\frac{7}{6} ln ∣ x + 7 ∣ - \frac{1}{6} ln ∣ x + 1 ∣) + C

Pasos de solución

\int \frac{33 x}{x ^{2} + 8 x + 7} d x

Sacar la constante:

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

= 33 \cdot \int \frac{x}{x ^{2} + 8 x + 7} d x

Completar el cuadrado

x^{2} + 8 x + 7 : (x + 4)^{2} - 9

= 33 \cdot \int \frac{x}{( x + 4 ) ^{2} - 9} d x

Aplicar integración por sustitución

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

= 33 \cdot \int \frac{u - 4}{u ^{2} - 9} d u

Tomar la fracción parcial de

\frac{u - 4}{u ^{2} - 9} : \frac{7}{6 ( u + 3 )} - \frac{1}{6 ( u - 3 )}

= 33 \cdot \int \frac{7}{6 ( u + 3 )} - \frac{1}{6 ( u - 3 )} d u

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

= 33 (\int \frac{7}{6 ( u + 3 )} d u - \int \frac{1}{6 ( u - 3 )} d u)

\int \frac{7}{6 ( u + 3 )} d u = \frac{7}{6} ln ∣ u + 3 ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

= 33 (\frac{7}{6} ln ∣ u + 3 ∣ - \frac{1}{6} ln ∣ u - 3 ∣)

Sustituir en la ecuación

u = x + 4

= 33 (\frac{7}{6} ln ∣ x + 4 + 3 ∣ - \frac{1}{6} ln ∣ x + 4 - 3 ∣)

Simplificar

33 (\frac{7}{6} ln ∣ x + 4 + 3 ∣ - \frac{1}{6} ln ∣ x + 4 - 3 ∣) : 33 (\frac{7}{6} ln ∣ x + 7 ∣ - \frac{1}{6} ln ∣ x + 1 ∣)

= 33 (\frac{7}{6} ln ∣ x + 7 ∣ - \frac{1}{6} ln ∣ x + 1 ∣)

Agregar una constante a la solución

= 33 (\frac{7}{6} ln ∣ x + 7 ∣ - \frac{1}{6} ln ∣ x + 1 ∣) + C

\int \frac{( e ^{2 y} - y )}{e ^{y}} d y

\frac{1}{s ^{2} ( s ^{2} + 4 )}

(x^{2} - 12) (2 x - 9)

\int_{x}^{2} 0.5 d y