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人気のある微分積分 >

integral of (a^2x+2a)/(a^2x^2-3ax-4)

解

\int \frac{a ^{2} x + 2 a}{a ^{2} x ^{2} - 3 a x - 4} d x

解

2 (\frac{1}{4} ln 4 a^{2} x^{2} - 12 a x - 16 - \frac{7}{20} (ln \frac{2}{5} + \frac{2}{5} a x - ln - \frac{8}{5} + \frac{2}{5} a x)) + C

解答ステップ

\int \frac{a ^{2} x + 2 a}{a ^{2} x ^{2} - 3 a x - 4} d x

平方完成する

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

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

u置換積分法を適用する

= \int \frac{2 ( 2 u + 7 )}{4 u ^{2} - 25} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{2 u + 7}{4 u ^{2} - 25} d u

拡張

\frac{2 u + 7}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} + \frac{7}{4 u ^{2} - 25}

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

= 2 (\int \frac{2 u}{4 u ^{2} - 25} d u + \int \frac{7}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{7}{4 u ^{2} - 25} d u = - \frac{7}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 25 - \frac{7}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1))

代用を戻す

u = x a^{2} - \frac{3 a}{2 a ^{2}}

= 2 (\frac{1}{4} ln 4 (x a^{2} - \frac{3 a}{2 a ^{2}})^{2} - 25 - \frac{7}{20} (ln \frac{2}{5} (x a^{2} - \frac{3 a}{2 a ^{2}}) + 1 - ln \frac{2}{5} (x a^{2} - \frac{3 a}{2 a ^{2}}) - 1))

簡素化

2 (\frac{1}{4} ln 4 (x a^{2} - \frac{3 a}{2 a ^{2}})^{2} - 25 - \frac{7}{20} (ln \frac{2}{5} (x a^{2} - \frac{3 a}{2 a ^{2}}) + 1 - ln \frac{2}{5} (x a^{2} - \frac{3 a}{2 a ^{2}}) - 1)) : 2 (\frac{1}{4} ln 4 a^{2} x^{2} - 12 a x - 16 - \frac{7}{20} (ln \frac{2}{5} + \frac{2}{5} a x - ln - \frac{8}{5} + \frac{2}{5} a x))

= 2 (\frac{1}{4} ln 4 a^{2} x^{2} - 12 a x - 16 - \frac{7}{20} (ln \frac{2}{5} + \frac{2}{5} a x - ln - \frac{8}{5} + \frac{2}{5} a x))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 a^{2} x^{2} - 12 a x - 16 - \frac{7}{20} (ln \frac{2}{5} + \frac{2}{5} a x - ln - \frac{8}{5} + \frac{2}{5} a x)) + C

グラフ

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