integral of (sqrt(2x)+1)^3\sqrt[4]{5/x}

解

\int (2 x + 1)^{3} 4 \frac{5}{x} d x

\frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{5 ^{\frac{3}{4}} 4 x} + \frac{1}{63 \cdot 5 ^{\frac{3}{4}}} (282 x^{\frac{9}{4}} + 135 x^{\frac{7}{4}} + 1262 x^{\frac{5}{4}} + 105 x^{\frac{3}{4}}) + C

解答ステップ

\int (2 x + 1)^{3} 4 \frac{5}{x} d x

部分積分を適用する

= \frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{5 ^{\frac{3}{4}} 4 x} - \int - \frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{4 \cdot 5 ^{\frac{3}{4}} x ^{\frac{5}{4}}} d x

\int - \frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{4 \cdot 5 ^{\frac{3}{4}} x ^{\frac{5}{4}}} d x = - \frac{1}{63 \cdot 5 ^{\frac{3}{4}}} (282 x^{\frac{9}{4}} + 135 x^{\frac{7}{4}} + 1262 x^{\frac{5}{4}} + 105 x^{\frac{3}{4}})

= \frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{5 ^{\frac{3}{4}} 4 x} - (- \frac{1}{63 \cdot 5 ^{\frac{3}{4}}} (282 x^{\frac{9}{4}} + 135 x^{\frac{7}{4}} + 1262 x^{\frac{5}{4}} + 105 x^{\frac{3}{4}}))

簡素化

= \frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{5 ^{\frac{3}{4}} 4 x} + \frac{1}{63 \cdot 5 ^{\frac{3}{4}}} (282 x^{\frac{9}{4}} + 135 x^{\frac{7}{4}} + 1262 x^{\frac{5}{4}} + 105 x^{\frac{3}{4}})

定数を解答に追加する

= \frac{4 2 x ^{\frac{5}{2}} + 15 x ^{2} + 10 2 x ^{\frac{3}{2}} + 5 x}{5 ^{\frac{3}{4}} 4 x} + \frac{1}{63 \cdot 5 ^{\frac{3}{4}}} (282 x^{\frac{9}{4}} + 135 x^{\frac{7}{4}} + 1262 x^{\frac{5}{4}} + 105 x^{\frac{3}{4}}) + C