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人気のある微分積分 >

integral of 1/(x^2sqrt(10x+1))

解

\int \frac{1}{x ^{2} 10 x + 1} d x

解

20 (\frac{1}{4} ln 10 x + 1 + 1 - \frac{1}{4 ( 10 x + 1 + 1 )} - \frac{1}{4} ln 10 x + 1 - 1 - \frac{1}{4 ( 10 x + 1 - 1 )}) + C

解答ステップ

\int \frac{1}{x ^{2} 10 x + 1} d x

u置換積分法を適用する

= \int \frac{20}{( u ^{2} - 1 ) ^{2}} d u

定数を除く:

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

= 20 \cdot \int \frac{1}{( u ^{2} - 1 ) ^{2}} d u

(u^{2} - 1)^{2} = u^{4} - 2 u^{2} + 1

= 20 \cdot \int \frac{1}{u ^{4} - 2 u ^{2} + 1} d u

以下の部分分数を得る:

\frac{1}{u ^{4} - 2 u ^{2} + 1} : \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}}

= 20 \cdot \int \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} - \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}} d u

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

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u + 1 ) ^{2}} d u = - \frac{1}{4 ( u + 1 )}

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

\int \frac{1}{4 ( u - 1 ) ^{2}} d u = - \frac{1}{4 ( u - 1 )}

= 20 (\frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4 ( u + 1 )} - \frac{1}{4} ln ∣ u - 1 ∣ - \frac{1}{4 ( u - 1 )})

代用を戻す

u = 10 x + 1

= 20 (\frac{1}{4} ln 10 x + 1 + 1 - \frac{1}{4 ( 10 x + 1 + 1 )} - \frac{1}{4} ln 10 x + 1 - 1 - \frac{1}{4 ( 10 x + 1 - 1 )})

定数を解答に追加する

= 20 (\frac{1}{4} ln 10 x + 1 + 1 - \frac{1}{4 ( 10 x + 1 + 1 )} - \frac{1}{4} ln 10 x + 1 - 1 - \frac{1}{4 ( 10 x + 1 - 1 )}) + C

グラフ

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