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人気のある微分積分 >

integral of 1/(x(x^{12)+1)^2}

解

\int \frac{1}{x ( x ^{12} + 1 ) ^{2}} d x

解

ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )} + C

解答ステップ

\int \frac{1}{x ( x ^{12} + 1 ) ^{2}} d x

u置換積分法を適用する

= \int \frac{1}{12 u ( u + 1 ) ^{2}} d u

定数を除く:

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

= \frac{1}{12} \cdot \int \frac{1}{u ( u + 1 ) ^{2}} d u

u置換積分法を適用する

= \frac{1}{12} \cdot \int \frac{1}{( v - 1 ) v ^{2}} d v

以下の部分分数を得る:

\frac{1}{( v - 1 ) v ^{2}} : \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}}

= \frac{1}{12} \cdot \int \frac{1}{v - 1} - \frac{1}{v} - \frac{1}{v ^{2}} d v

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

= \frac{1}{12} (\int \frac{1}{v - 1} d v - \int \frac{1}{v} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

= \frac{1}{12} (ln ∣ v - 1 ∣ - ln ∣ v ∣ - (- \frac{1}{v}))

代用を戻す

= \frac{1}{12} (ln x^{12} + 1 - 1 - ln x^{12} + 1 - (- \frac{1}{x ^{12} + 1}))

簡素化

\frac{1}{12} (ln x^{12} + 1 - 1 - ln x^{12} + 1 - (- \frac{1}{x ^{12} + 1})) : ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )}

= ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )}

定数を解答に追加する

= ln (x) - \frac{1}{12} ln x^{12} + 1 + \frac{1}{12 ( x ^{12} + 1 )} + C

グラフ

人気の例

integral of xln(1-x)

\int x ln (1 - x) d x

integral of ysqrt(y+7)

\int y y + 7 d y

integral of tan^3(x)+1

\int tan^{3} (x) + 1 d x

integral of 1/(sqrt(9-2x^2))

\int \frac{1}{9 - 2 x ^{2}} d x

integral of (3x^2+2x)

\int (3 x^{2} + 2 x) d x