integral from 1 to 3 of (5x+3)/(x^2+2x)

解

\int_{1}^{3} \frac{5 x + 3}{x ^{2} + 2 x} d x

\frac{- 4 ln ( 3 ) + 7 ln ( 5 )}{2}

十進法表記

3.43580 \dots

解答ステップ

\int_{1}^{3} \frac{5 x + 3}{x ^{2} + 2 x} d x

以下の部分分数を得る:

\frac{5 x + 3}{x ^{2} + 2 x} : \frac{3}{2 x} + \frac{7}{2 ( x + 2 )}

= \int_{1}^{3} \frac{3}{2 x} + \frac{7}{2 ( 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{3}{2 x} d x + \int_{1}^{3} \frac{7}{2 ( x + 2 )} d x

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

\int_{1}^{3} \frac{7}{2 ( x + 2 )} d x = \frac{7}{2} (ln (5) - ln (3))

= \frac{3}{2} ln (3) + \frac{7}{2} (ln (5) - ln (3))

簡素化

\frac{3}{2} ln (3) + \frac{7}{2} (ln (5) - ln (3)) : \frac{- 4 ln ( 3 ) + 7 ln ( 5 )}{2}

= \frac{- 4 ln ( 3 ) + 7 ln ( 5 )}{2}