integral of (6x+7)/((9x^2+21x)^3)

解

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{3}} d x

\frac{1}{3087 x} - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}} + C

解答ステップ

\int \frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{3}} d x

以下の部分分数を得る:

\frac{6 x + 7}{( 9 x ^{2} + 21 x ) ^{3}} : - \frac{1}{3087 x ^{2}} + \frac{1}{1323 x ^{3}} + \frac{1}{343 ( 3 x + 7 ) ^{2}} + \frac{1}{49 ( 3 x + 7 ) ^{3}}

= \int - \frac{1}{3087 x ^{2}} + \frac{1}{1323 x ^{3}} + \frac{1}{343 ( 3 x + 7 ) ^{2}} + \frac{1}{49 ( 3 x + 7 ) ^{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}{3087 x ^{2}} d x + \int \frac{1}{1323 x ^{3}} d x + \int \frac{1}{343 ( 3 x + 7 ) ^{2}} d x + \int \frac{1}{49 ( 3 x + 7 ) ^{3}} d x

\int \frac{1}{3087 x ^{2}} d x = - \frac{1}{3087 x}

\int \frac{1}{1323 x ^{3}} d x = - \frac{1}{2646 x ^{2}}

\int \frac{1}{343 ( 3 x + 7 ) ^{2}} d x = - \frac{1}{1029 ( 3 x + 7 )}

\int \frac{1}{49 ( 3 x + 7 ) ^{3}} d x = - \frac{1}{294 ( 3 x + 7 ) ^{2}}

= - (- \frac{1}{3087 x}) - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}}

簡素化

= \frac{1}{3087 x} - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}}

定数を解答に追加する

= \frac{1}{3087 x} - \frac{1}{2646 x ^{2}} - \frac{1}{1029 ( 3 x + 7 )} - \frac{1}{294 ( 3 x + 7 ) ^{2}} + C