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人気のある微分積分 >

integral of 1/(4a^2+12a+5)

解

\int \frac{1}{4 a ^{2} + 12 a + 5} d a

解

- \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2}) + C

解答ステップ

\int \frac{1}{4 a ^{2} + 12 a + 5} d a

平方完成する

4 a^{2} + 12 a + 5 : 4 (a + \frac{3}{2})^{2} - 4

= \int \frac{1}{4 ( a + \frac{3}{2} ) ^{2} - 4} d a

u置換積分法を適用する

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

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

= \int \frac{1}{4} \cdot \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

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

因数

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

= \frac{1}{4} \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

= \frac{1}{4} (- \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}

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

代用を戻す

u = a + \frac{3}{2}

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

簡素化

\frac{1}{4} (- (\frac{ln a + \frac{3}{2} + 1}{2} - \frac{ln a + \frac{3}{2} - 1}{2})) : - \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2})

= - \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2})

定数を解答に追加する

= - \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2}) + C

グラフ

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