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integral de+(16)/((64x^2+1)^2)

Solución

\int + \frac{16}{( 64 x ^{2} + 1 ) ^{2}} d x

arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)) + C

Pasos de solución

\int \frac{16}{( 64 x ^{2} + 1 ) ^{2}} d x

Sacar la constante:

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

= 16 \cdot \int \frac{1}{( 64 x ^{2} + 1 ) ^{2}} d x

Aplicar integración por sustitución

= 16 \cdot \int \frac{1}{8 ( u ^{2} + 1 ) ^{2}} d u

Sacar la constante:

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

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Aplicar integración por sustitución trigonométrica

: \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

= 16 \cdot \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( v ) cos ^{2} ( v )} d v

Re-escribir usando identidades trigonométricas

= 16 \cdot \frac{1}{8} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Sacar la constante:

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

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

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

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Sustitución hacia atrás

= 16 \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)))

Simplificar

16 \cdot \frac{1}{8} \cdot \frac{1}{2} (arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))) : arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))

= arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x))

Agregar una constante a la solución

= arctan (8 x) + \frac{1}{2} sin (2 arctan (8 x)) + C

\int_{- \infty}^{0} \frac{1}{4 - 5 x} d x

x \to - 1 lim (\frac{x + 3}{x ^{2} + 1})

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

\int \frac{1}{x ^{1.8}} d x

\frac{d y}{d x} - cos (x) y = \frac{1}{100} cos (x)