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人気のある微分積分 >

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

解

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

解

\frac{1}{4} (- - 4 x^{2} + 4 x + 3 + 7 arcsin (x - \frac{1}{2})) + C

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

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

拡張

\frac{2 u + 7}{- u ^{2} + 1} : \frac{2 u}{- u ^{2} + 1} + \frac{7}{- u ^{2} + 1}

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

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}{4} (\int \frac{2 u}{- u ^{2} + 1} d u + \int \frac{7}{- u ^{2} + 1} d u)

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

\int \frac{7}{- u ^{2} + 1} d u = 7 arcsin (u)

= \frac{1}{4} (- 2 - u^{2} + 1 + 7 arcsin (u))

代用を戻す

u = x - \frac{1}{2}

= \frac{1}{4} - 2 - (x - \frac{1}{2})^{2} + 1 + 7 arcsin (x - \frac{1}{2})

簡素化

\frac{1}{4} - 2 - (x - \frac{1}{2})^{2} + 1 + 7 arcsin (x - \frac{1}{2}) : \frac{1}{4} (- - 4 x^{2} + 4 x + 3 + 7 arcsin (x - \frac{1}{2}))

= \frac{1}{4} (- - 4 x^{2} + 4 x + 3 + 7 arcsin (x - \frac{1}{2}))

定数を解答に追加する

= \frac{1}{4} (- - 4 x^{2} + 4 x + 3 + 7 arcsin (x - \frac{1}{2})) + C

グラフ

人気の例

integral of (x^2+4)^{3/2}

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

integral of 1/(sqrt(5-2x-x^2))

\int \frac{1}{5 - 2 x - x ^{2}} d x

integral of (x-7)/((x+1)(x^2+2x+5))

\int \frac{x - 7}{( x + 1 ) ( x ^{2} + 2 x + 5 )} d x

integral of (cos(5sqrt(x)))/(sqrt(x))

\int \frac{cos ( 5 x )}{x} d x

integral of cos^2(sqrt(x))

\int cos^{2} (x) d x