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

解答

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

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

求解步骤

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

使用指数法则:

a^{- b} = \frac{1}{a ^{b}}

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

使用分布积分法

= \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}}})

化简

= ln x + x^{2} + 1

解答补常数

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

s l o p e (- \frac{2}{7}, \frac{2}{3}), (\frac{5}{7}, - \frac{7}{3})

x \to 1 lim (\frac{x - 1}{x ^{n} - 1})

(x + y)^{2} d x + (2 x y + x^{2} - 4) d y = 0

\frac{d}{d x} (5 e^{x} + 3)

\frac{d}{d x} ((x^{2} + 4) e^{- 2 x})