integral from 3 to 4 of (x^2-1)/(x-1)

解

\int_{3}^{4} \frac{x ^{2} - 1}{x - 1} d x

\frac{9}{2}

十進法表記

4.5

解答ステップ

\int_{3}^{4} \frac{x ^{2} - 1}{x - 1} d x

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

= \int_{3}^{4} \frac{( x ^{2} - 1 ) ( x + 1 )}{x ^{2} - 1 ^{2}} d x

簡素化

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

= \int_{3}^{4} x + 1 d x

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_{3}^{4} x d x + \int_{3}^{4} 1 d x

\int_{3}^{4} x d x = \frac{7}{2}

\int_{3}^{4} 1 d x = 1

= \frac{7}{2} + 1

簡素化

\frac{7}{2} + 1 : \frac{9}{2}

= \frac{9}{2}