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

Lösung

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

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

Schritte zur Lösung

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

Vereinfache

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

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

Wende die partielle Integration an

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

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

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

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