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

解

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

\frac{1}{5} ln (3)

十進法表記

0.21972 \dots

解答ステップ

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

以下の部分分数を得る:

\frac{1}{2 x ^{2} + 3 x - 2} : \frac{2}{5 ( 2 x - 1 )} - \frac{1}{5 ( x + 2 )}

= \int_{1}^{3} \frac{2}{5 ( 2 x - 1 )} - \frac{1}{5 ( x + 2 )} 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_{1}^{3} \frac{2}{5 ( 2 x - 1 )} d x - \int_{1}^{3} \frac{1}{5 ( x + 2 )} d x

\int_{1}^{3} \frac{2}{5 ( 2 x - 1 )} d x = \frac{1}{5} ln (5)

\int_{1}^{3} \frac{1}{5 ( x + 2 )} d x = \frac{1}{5} (ln (5) - ln (3))

= \frac{1}{5} ln (5) - \frac{1}{5} (ln (5) - ln (3))

簡素化

\frac{1}{5} ln (5) - \frac{1}{5} (ln (5) - ln (3)) : \frac{1}{5} ln (3)

= \frac{1}{5} ln (3)