integral of (x^2+2)/(sqrt(1-x^2))

Lösung

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

\frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

Schritte zur Lösung

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

Multipliziere aus

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

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

Wende die Summenregel an:

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

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

\int \frac{x ^{2}}{1 - x ^{2}} d x = \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

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

= \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + 2 arcsin (x)

Vereinfache

\frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + 2 arcsin (x) : \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

= \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x))

Füge eine Konstante zur Lösung hinzu

= \frac{5}{2} arcsin (x) - \frac{1}{4} sin (2 arcsin (x)) + C

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

\int x ln (9 + x^{2}) d x

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

\int \frac{7}{x ( ln ( x ) + 1 )} d x

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