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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

= \int \frac{u}{2 u ^{2} - 8 u + 20} d u

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{u}{u ^{2} - 8 u + 20} d u

平方完成する

u^{2} - 8 u + 20 : (u - 4)^{2} + 4

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

u置換積分法を適用する

= \frac{1}{2} \cdot \int \frac{v + 4}{v ^{2} + 4} d v

拡張

\frac{v + 4}{v ^{2} + 4} : \frac{v}{v ^{2} + 4} + \frac{4}{v ^{2} + 4}

= \frac{1}{2} \cdot \int \frac{v}{v ^{2} + 4} + \frac{4}{v ^{2} + 4} 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

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

\int \frac{v}{v ^{2} + 4} d v = v^{2} + 4

\int \frac{4}{v ^{2} + 4} d v = 4 ln (\frac{1}{2} v + 4 + v^{2})

= \frac{1}{2} (v^{2} + 4 + 4 ln (\frac{1}{2} v + 4 + v^{2}))

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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

\int \frac{5}{x ( - 1 + ln ( - 2 x ) ) ^{3}} d x

integral of 7/(\sqrt[3]{x)}

\int \frac{7}{3 x} d x

integral of sqrt(25)

\int 25 d x

integral of 6+tan^2(x)

\int 6 + tan^{2} (x) d x

integral of ((tan(x))/(cos(x))-csc^2(x))

\int (\frac{tan ( x )}{cos ( x )} - csc^{2} (x)) d x