integral of (1-4x^2)^{3/2}

Lösung

\int (1 - 4 x^{2})^{\frac{3}{2}} d x

x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x)) + C

Schritte zur Lösung

\int (1 - 4 x^{2})^{\frac{3}{2}} d x

Wende integrale Substitution an

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int (1 - u^{2})^{\frac{3}{2}} d u

Wende die partielle Integration an

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

\int - 3 u^{2} (1 - u^{2})^{\frac{1}{2}} d u = - \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

= \frac{1}{2} (u (1 - u^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))))

Setze in

u = 2 x

ein

= \frac{1}{2} (2 x (1 - (2 x)^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (2 x) - \frac{1}{4} sin (4 arcsin (2 x)))))

Vereinfache

\frac{1}{2} (2 x (1 - (2 x)^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (2 x) - \frac{1}{4} sin (4 arcsin (2 x))))) : x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x))

= x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x))

Füge eine Konstante zur Lösung hinzu

= x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x)) + C

\int \frac{5 x ^{2} - 2 x + 64}{x ^{3} - 16 x} d x

\int (e^{2 x} + 1)^{10} e^{2 x} d x

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

\int 6 x (3 x^{2} - 1) d x

in t e g r a l \frac{1}{x ^{2} + 9}