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人気のある微分積分 >

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

解

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

解

\frac{1}{4} (5 (ln 2 x + 2 \frac{4 x ^{2} + 4 x + 3}{2} + 1 - \frac{ln ( 2 )}{2}) + 4 x^{2} + 4 x + 3) + C

解答ステップ

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

x + 3 = 3 (\frac{x}{3} + 1)

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

u置換積分法を適用する

= 3 \cdot \int \frac{3 u}{36 u ^{2} - 60 u + 27} d u

定数を除く:

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

= 3 \cdot 3 \cdot \int \frac{u}{36 u ^{2} - 60 u + 27} d u

平方完成する

36 u^{2} - 60 u + 27 : 36 (u - \frac{5}{6})^{2} + 2

= 3 \cdot 3 \cdot \int \frac{u}{36 ( u - \frac{5}{6} ) ^{2} + 2} d u

u置換積分法を適用する

= 3 \cdot 3 \cdot \int \frac{6 v + 5}{6 36 v ^{2} + 2} d v

定数を除く:

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

= 3 \cdot 3 \cdot \frac{1}{6} \cdot \int \frac{6 v + 5}{36 v ^{2} + 2} d v

拡張

\frac{6 v + 5}{36 v ^{2} + 2} : \frac{6 v}{36 v ^{2} + 2} + \frac{5}{36 v ^{2} + 2}

= 3 \cdot 3 \cdot \frac{1}{6} \cdot \int \frac{6 v}{36 v ^{2} + 2} + \frac{5}{36 v ^{2} + 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

= 3 \cdot 3 \cdot \frac{1}{6} (\int \frac{6 v}{36 v ^{2} + 2} d v + \int \frac{5}{36 v ^{2} + 2} d v)

\int \frac{6 v}{36 v ^{2} + 2} d v = \frac{1}{6} 36 v^{2} + 2

\int \frac{5}{36 v ^{2} + 2} d v = \frac{5}{6} ln 32 v + 1 + 18 v^{2}

= 3 \cdot 3 \cdot \frac{1}{6} (\frac{1}{6} 36 v^{2} + 2 + \frac{5}{6} ln 32 v + 1 + 18 v^{2})

代用を戻す

= 3 \cdot 3 \cdot \frac{1}{6} \frac{1}{6} 36 (\frac{x}{3} + 1 - \frac{5}{6})^{2} + 2 + \frac{5}{6} ln 32 (\frac{x}{3} + 1 - \frac{5}{6}) + 1 + 18 (\frac{x}{3} + 1 - \frac{5}{6})^{2}

簡素化

3 \cdot 3 \cdot \frac{1}{6} \frac{1}{6} 36 (\frac{x}{3} + 1 - \frac{5}{6})^{2} + 2 + \frac{5}{6} ln 32 (\frac{x}{3} + 1 - \frac{5}{6}) + 1 + 18 (\frac{x}{3} + 1 - \frac{5}{6})^{2} : \frac{1}{4} (5 (ln 2 x + 2 \frac{4 x ^{2} + 4 x + 3}{2} + 1 - \frac{ln ( 2 )}{2}) + 4 x^{2} + 4 x + 3)

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

定数を解答に追加する

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

グラフ

人気の例

integral of 1/((y+1)^2)

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

integral of (6+9x)/(1+x^2)

\int \frac{6 + 9 x}{1 + x ^{2}} d x

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

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

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

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

integral of 1/((x+1)^{2/3)}

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