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人気のある微分積分 >

integral of ((4x-5))/(x^2+8x+7)

解

\int \frac{( 4 x - 5 )}{x ^{2} + 8 x + 7} d x

解

2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

解答ステップ

\int \frac{4 x - 5}{x ^{2} + 8 x + 7} d x

平方完成する

x^{2} + 8 x + 7 : (x + 4)^{2} - 9

= \int \frac{4 x - 5}{( x + 4 ) ^{2} - 9} d x

u置換積分法を適用する

= \int \frac{4 u - 21}{u ^{2} - 9} d u

因数

\frac{4 u - 21}{u ^{2} - 9}

= \int \frac{4 u - 21}{( u + 3 ) ( u - 3 )} d u

拡張

\frac{4 u - 21}{( u + 3 ) ( u - 3 )} : \frac{4 u}{( u + 3 ) ( u - 3 )} - \frac{21}{( u + 3 ) ( u - 3 )}

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{4 u}{( u + 3 ) ( u - 3 )} d u - \int \frac{21}{( u + 3 ) ( u - 3 )} d u

\int \frac{4 u}{( u + 3 ) ( u - 3 )} d u = 2 ln (∣ u + 3 ∣ ∣ u - 3 ∣)

\int \frac{21}{( u + 3 ) ( u - 3 )} d u = 21 (- \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣)

= 2 ln (∣ u + 3 ∣ ∣ u - 3 ∣) - 21 (- \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣)

代用を戻す

u = x + 4

= 2 ln (∣ x + 4 + 3 ∣ ∣ x + 4 - 3 ∣) - 21 (- \frac{1}{6} ln ∣ x + 4 + 3 ∣ + \frac{1}{6} ln ∣ x + 4 - 3 ∣)

簡素化

2 ln (∣ x + 4 + 3 ∣ ∣ x + 4 - 3 ∣) - 21 (- \frac{1}{6} ln ∣ x + 4 + 3 ∣ + \frac{1}{6} ln ∣ x + 4 - 3 ∣) : 2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

= 2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

定数を解答に追加する

= 2 ln (∣ x + 7 ∣ ∣ x + 1 ∣) - 21 (- \frac{1}{6} ln ∣ x + 7 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

グラフ

人気の例

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