integral of 1/(sqrt(x+1)+\sqrt{x-1)}

解

\int \frac{1}{x + 1 + x - 1} d x

\frac{1}{3} (x + 1)^{\frac{3}{2}} - \frac{1}{3} (x - 1)^{\frac{3}{2}} + C

解答ステップ

\int \frac{1}{x + 1 + x - 1} d x

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

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

拡張

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

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

定数を解答に追加する

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