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人気のある微分積分 >

integral of (5x)/(x^2+3x+4)

解

\int \frac{5 x}{x ^{2} + 3 x + 4} d x

解

10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7})) + C

解答ステップ

\int \frac{5 x}{x ^{2} + 3 x + 4} d x

定数を除く:

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

= 5 \cdot \int \frac{x}{x ^{2} + 3 x + 4} d x

平方完成する

x^{2} + 3 x + 4 : (x + \frac{3}{2})^{2} + \frac{7}{4}

= 5 \cdot \int \frac{x}{( x + \frac{3}{2} ) ^{2} + \frac{7}{4}} d x

u置換積分法を適用する

= 5 \cdot \int \frac{2 ( 2 u - 3 )}{4 u ^{2} + 7} d u

定数を除く:

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

= 5 \cdot 2 \cdot \int \frac{2 u - 3}{4 u ^{2} + 7} d u

拡張

\frac{2 u - 3}{4 u ^{2} + 7} : \frac{2 u}{4 u ^{2} + 7} - \frac{3}{4 u ^{2} + 7}

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

= 5 \cdot 2 (\int \frac{2 u}{4 u ^{2} + 7} d u - \int \frac{3}{4 u ^{2} + 7} d u)

\int \frac{2 u}{4 u ^{2} + 7} d u = \frac{1}{4} ln 4 u^{2} + 7

\int \frac{3}{4 u ^{2} + 7} d u = \frac{3}{2 7} arctan (\frac{2}{7} u)

= 5 \cdot 2 (\frac{1}{4} ln 4 u^{2} + 7 - \frac{3}{2 7} arctan (\frac{2}{7} u))

代用を戻す

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

= 5 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 7 - \frac{3}{2 7} arctan (\frac{2}{7} (x + \frac{3}{2})))

簡素化

5 \cdot 2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 7 - \frac{3}{2 7} arctan (\frac{2}{7} (x + \frac{3}{2}))) : 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7}))

= 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7}))

定数を解答に追加する

= 10 (\frac{1}{4} ln 4 x^{2} + 12 x + 16 - \frac{3}{2 7} arctan (\frac{2 x + 3}{7})) + C

グラフ

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