integral of (x^5)/((2x^2+3)^3)

解

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

\frac{1}{16} ln 2 x^{2} + 3 + \frac{3}{8 ( 2 x ^{2} + 3 )} - \frac{9}{32 ( 2 x ^{2} + 3 ) ^{2}} + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{5}}{( 2 x ^{2} + 3 ) ^{3}} : \frac{x}{4 ( 2 x ^{2} + 3 )} - \frac{3 x}{2 ( 2 x ^{2} + 3 ) ^{2}} + \frac{9 x}{4 ( 2 x ^{2} + 3 ) ^{3}}

= \int \frac{x}{4 ( 2 x ^{2} + 3 )} - \frac{3 x}{2 ( 2 x ^{2} + 3 ) ^{2}} + \frac{9 x}{4 ( 2 x ^{2} + 3 ) ^{3}} 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 \frac{x}{4 ( 2 x ^{2} + 3 )} d x - \int \frac{3 x}{2 ( 2 x ^{2} + 3 ) ^{2}} d x + \int \frac{9 x}{4 ( 2 x ^{2} + 3 ) ^{3}} d x

\int \frac{x}{4 ( 2 x ^{2} + 3 )} d x = \frac{1}{16} ln 2 x^{2} + 3

\int \frac{3 x}{2 ( 2 x ^{2} + 3 ) ^{2}} d x = - \frac{3}{8 ( 2 x ^{2} + 3 )}

\int \frac{9 x}{4 ( 2 x ^{2} + 3 ) ^{3}} d x = - \frac{9}{32 ( 2 x ^{2} + 3 ) ^{2}}

= \frac{1}{16} ln 2 x^{2} + 3 - (- \frac{3}{8 ( 2 x ^{2} + 3 )}) - \frac{9}{32 ( 2 x ^{2} + 3 ) ^{2}}

簡素化

= \frac{1}{16} ln 2 x^{2} + 3 + \frac{3}{8 ( 2 x ^{2} + 3 )} - \frac{9}{32 ( 2 x ^{2} + 3 ) ^{2}}

定数を解答に追加する

= \frac{1}{16} ln 2 x^{2} + 3 + \frac{3}{8 ( 2 x ^{2} + 3 )} - \frac{9}{32 ( 2 x ^{2} + 3 ) ^{2}} + C