integral de (x^3)/(x^4sqrt(16-x^2))

Solución

\int \frac{x ^{3}}{x ^{4} 16 - x ^{2}} d x

- \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1) + C

Pasos de solución

\int \frac{x ^{3}}{x ^{4} 16 - x ^{2}} d x

Simplificar

\frac{x ^{3}}{x ^{4} 16 - x ^{2}} : \frac{1}{x - x ^{2} + 16}

= \int \frac{1}{x - x ^{2} + 16} d x

Aplicar integración por sustitución

: \int - \frac{1}{- u ^{2} + 16} d u

= \int - \frac{1}{- u ^{2} + 16} d u

Sacar la constante:

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

= - \int \frac{1}{- u ^{2} + 16} d u

Aplicar integración por sustitución

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

Sacar la constante:

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

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

Aplicar la regla de integración:

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

= - \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Sustitución hacia atrás

= - \frac{1}{4} \frac{ln \frac{- x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- x ^{2} + 16}{4} - 1}{2}

Simplificar

- \frac{1}{4} \frac{ln \frac{- x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- x ^{2} + 16}{4} - 1}{2} : - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1)

= - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1)

Agregar una constante a la solución

= - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1) + C

\int (4 x^{2} - 12 x + 9)^{3} d x

\int (x^{2} - \frac{1}{x ^{2}})^{3} x d x

\int x 1 - 3 x^{2} d x

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

\int (cos (2 x)) (e^{s i n (2 x)}) d x