integral from 1 to 5 of (4x^3-1)/(2x-1)

解

\int_{1}^{5} \frac{4 x ^{3} - 1}{2 x - 1} d x

\frac{290}{3} - \frac{1}{2} ln (3)

十進法表記

96.11736 \dots

解答ステップ

\int_{1}^{5} \frac{4 x ^{3} - 1}{2 x - 1} d x

長除法

\frac{4 x ^{3} - 1}{2 x - 1} : 2 x^{2} + x + \frac{1}{2} - \frac{1}{2 ( 2 x - 1 )}

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

\int_{1}^{5} 2 x^{2} d x = \frac{248}{3}

\int_{1}^{5} x d x = 12

\int_{1}^{5} \frac{1}{2} d x = 2

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

= \frac{248}{3} + 12 + 2 - \frac{1}{2} ln (3)

簡素化

\frac{248}{3} + 12 + 2 - \frac{1}{2} ln (3) : \frac{290}{3} - \frac{1}{2} ln (3)

= \frac{290}{3} - \frac{1}{2} ln (3)