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

Lösung

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

ln x^{2} + 1 + x + C

Schritte zur Lösung

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

\frac{( 1 + x ) ^{2}}{x ^{2} + 1} = \frac{2 x}{x ^{2} + 1} + 1

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

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

\int 1 d x = x

= ln x^{2} + 1 + x

Füge eine Konstante zur Lösung hinzu

= ln x^{2} + 1 + x + C

d er i v a t i v e \frac{( 8 u ^{2} )}{( u ^{2} + u ) ^{3}}

t an g e n t f (x) = x^{2} + x, at x = - 1

\frac{d}{d x} (\frac{2901}{1 + 4.76 e ^{- 0.3 x}})

\frac{\partial}{\partial y} (sin^{2} (m x + n y))

\frac{d}{d x} (ln (x^{5} - 3 x^{3}))