integral of (8x+6)/((4x^2+6x)^3)

解

\int \frac{8 x + 6}{( 4 x ^{2} + 6 x ) ^{3}} d x

\frac{1}{54 x} - \frac{1}{72 x ^{2}} - \frac{1}{27 ( 2 x + 3 )} - \frac{1}{18 ( 2 x + 3 ) ^{2}} + C

解答ステップ

\int \frac{8 x + 6}{( 4 x ^{2} + 6 x ) ^{3}} d x

以下の部分分数を得る:

\frac{8 x + 6}{( 4 x ^{2} + 6 x ) ^{3}} : - \frac{1}{54 x ^{2}} + \frac{1}{36 x ^{3}} + \frac{2}{27 ( 2 x + 3 ) ^{2}} + \frac{2}{9 ( 2 x + 3 ) ^{3}}

= \int - \frac{1}{54 x ^{2}} + \frac{1}{36 x ^{3}} + \frac{2}{27 ( 2 x + 3 ) ^{2}} + \frac{2}{9 ( 2 x + 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{1}{54 x ^{2}} d x + \int \frac{1}{36 x ^{3}} d x + \int \frac{2}{27 ( 2 x + 3 ) ^{2}} d x + \int \frac{2}{9 ( 2 x + 3 ) ^{3}} d x

\int \frac{1}{54 x ^{2}} d x = - \frac{1}{54 x}

\int \frac{1}{36 x ^{3}} d x = - \frac{1}{72 x ^{2}}

\int \frac{2}{27 ( 2 x + 3 ) ^{2}} d x = - \frac{1}{27 ( 2 x + 3 )}

\int \frac{2}{9 ( 2 x + 3 ) ^{3}} d x = - \frac{1}{18 ( 2 x + 3 ) ^{2}}

= - (- \frac{1}{54 x}) - \frac{1}{72 x ^{2}} - \frac{1}{27 ( 2 x + 3 )} - \frac{1}{18 ( 2 x + 3 ) ^{2}}

簡素化

= \frac{1}{54 x} - \frac{1}{72 x ^{2}} - \frac{1}{27 ( 2 x + 3 )} - \frac{1}{18 ( 2 x + 3 ) ^{2}}

定数を解答に追加する

= \frac{1}{54 x} - \frac{1}{72 x ^{2}} - \frac{1}{27 ( 2 x + 3 )} - \frac{1}{18 ( 2 x + 3 ) ^{2}} + C