integral of (x^2+3)/(x^2-4)

Lösung

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

- \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x + C

Schritte zur Lösung

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

\frac{x ^{2} + 3}{x ^{2} - 4} = \frac{7}{x ^{2} - 4} + 1

= \int \frac{7}{x ^{2} - 4} + 1 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{7}{x ^{2} - 4} d x + \int 1 d x

\int \frac{7}{x ^{2} - 4} d x = - \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

\int 1 d x = x

= - \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x

Füge eine Konstante zur Lösung hinzu

= - \frac{7}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + x + C

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

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

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

in t e g r a l sec^{3} (x)

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