integral 1/((1+x^2)^{0.5)}

Lösung

integral

\frac{1}{( 1 + x ^{2} ) ^{0.5}}

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

Schritte zur Lösung

\int \frac{1}{( 1 + x ^{2} ) ^{0.5}} d x

Vereinfache

x^{2} : xx

= \int \frac{1}{( 1 + xx ) ^{0.5}} d x

xx = x^{2}

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

Wende die partielle Integration an

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

\int - \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x = - ln x + 1 + x^{2} + \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

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

Vereinfache

= ln x + x^{2} + 1

Füge eine Konstante zur Lösung hinzu

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

in t e g r a l \frac{- 1}{y + 1}

in t e g r a l x^{2} e^{2} x

in t e g r a l \frac{2 x}{( x + y ) ^{3}}

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

in t e g r a l e^{- x} cos^{2} (x)