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人気のある微分積分 >

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

解

\int \frac{3 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )} d x

解

- \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

解答ステップ

\int \frac{3 x}{( x ^{2} - 1 ) ( x ^{2} + 1 )} d x

定数を除く:

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

= 3 \cdot \int \frac{x}{( x ^{2} - 1 ) ( x ^{2} + 1 )} d x

u置換積分法を適用する

= 3 \cdot \int \frac{1}{2 ( u - 1 ) ( u + 1 )} d u

定数を除く:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{( u - 1 ) ( u + 1 )} d u

拡張

(u - 1) (u + 1) : u^{2} - 1

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} - 1} d u

因数

u^{2} - 1 : - (- u^{2} + 1)

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

定数を除く:

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

= 3 \cdot \frac{1}{2} (- \int \frac{1}{- u ^{2} + 1} d u)

共通積分を使用する:

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

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

代用を戻す

u = x^{2}

= 3 \cdot \frac{1}{2} (- (\frac{ln x ^{2} + 1}{2} - \frac{ln x ^{2} - 1}{2}))

簡素化

3 \cdot \frac{1}{2} (- (\frac{ln x ^{2} + 1}{2} - \frac{ln x ^{2} - 1}{2})) : - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1)

= - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1)

定数を解答に追加する

= - \frac{3}{4} (ln x^{2} + 1 - ln x^{2} - 1) + C

グラフ

人気の例

integral of (3x+1)e^{-x}

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

integral of (sin^2(ln(x)))/(xcos(ln(x)))

\int \frac{sin ^{2} ( ln ( x ) )}{x cos ( ln ( x ) )} d x

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

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

integral of x/((1-cx))

\int \frac{x}{( 1 - c x )} d x

integral of [te^t]

\int [t e^{t}] d t