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人気のある微分積分 >

integral of 4/(x(x^2+1)^2)

解

\int \frac{4}{x ( x ^{2} + 1 ) ^{2}} d x

解

2 (2 ln (x) - ln x^{2} + 1 + \frac{1}{x ^{2} + 1}) + C

解答ステップ

\int \frac{4}{x ( x ^{2} + 1 ) ^{2}} d x

定数を除く:

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

= 4 \cdot \int \frac{1}{x ( x ^{2} + 1 ) ^{2}} d x

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

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

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 4 \cdot \frac{1}{2} (ln ∣ u - 1 ∣ - ln ∣ u ∣ - (- \frac{1}{u}))

代用を戻す

u = x^{2} + 1

= 4 \cdot \frac{1}{2} (ln x^{2} + 1 - 1 - ln x^{2} + 1 - (- \frac{1}{x ^{2} + 1}))

簡素化

4 \cdot \frac{1}{2} (ln x^{2} + 1 - 1 - ln x^{2} + 1 - (- \frac{1}{x ^{2} + 1})) : 2 (2 ln (x) - ln x^{2} + 1 + \frac{1}{x ^{2} + 1})

= 2 (2 ln (x) - ln x^{2} + 1 + \frac{1}{x ^{2} + 1})

定数を解答に追加する

= 2 (2 ln (x) - ln x^{2} + 1 + \frac{1}{x ^{2} + 1}) + C

グラフ

人気の例

integral of x^2sqrt((4-x^2))

\int x^{2} (4 - x^{2}) d x

integral of 12x^2-4x

\int 12 x^{2} - 4 x d x

integral of-xe^{4x}

\int - x e^{4 x} d x

integral of 1/(3+5x^2)

\int \frac{1}{3 + 5 x ^{2}} d x

integral of 1/(sqrt(t)+1)

\int \frac{1}{t + 1} d t