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人気のある微分積分 >

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

解

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

解

\frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1) + C

解答ステップ

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

u置換積分法を適用する

= \int \frac{3 u + 13}{2 2 u ( u + 1 )} d u

定数を除く:

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

= \frac{1}{2 2} \cdot \int \frac{3 u + 13}{u ( u + 1 )} d u

拡張

\frac{3 u + 13}{u ( u + 1 )} : \frac{3 u}{u ^{2} + u} + \frac{13}{u ^{2} + u}

= \frac{1}{2 2} \cdot \int \frac{3 u}{u ^{2} + u} + \frac{13}{u ^{2} + u} 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}{2 2} (\int \frac{3 u}{u ^{2} + u} d u + \int \frac{13}{u ^{2} + u} d u)

\int \frac{3 u}{u ^{2} + u} d u = 3 (\frac{u}{u + 1} (u + 1) - ln u + 1 + u)

\int \frac{13}{u ^{2} + u} d u = 26 ln u + 1 + u

= \frac{1}{2 2} (3 (\frac{u}{u + 1} (u + 1) - ln u + 1 + u) + 26 ln u + 1 + u)

代用を戻す

u = 2 x - 1

= \frac{1}{2 2} (3 (\frac{2 x - 1}{2 x - 1 + 1} (2 x - 1 + 1) - ln 2 x - 1 + 1 + 2 x - 1) + 26 ln 2 x - 1 + 1 + 2 x - 1)

簡素化

\frac{1}{2 2} (3 (\frac{2 x - 1}{2 x - 1 + 1} (2 x - 1 + 1) - ln 2 x - 1 + 1 + 2 x - 1) + 26 ln 2 x - 1 + 1 + 2 x - 1) : \frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1)

= \frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1)

定数を解答に追加する

= \frac{1}{2 2} (3 \frac{2 ( 2 x - 1 )}{x} x + 23 ln 2 x + 2 x - 1) + C

グラフ

人気の例

derivative t^3e^tt

d er i v a t i v e t^{3} e^{t} t

limit as n approaches infinity of c^n

n \to \infty lim (c^{n})

y^{''}+y^'-2y=0,y(0)=1,y^'(0)=1

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1

(3 cos (t))^{^{'}}

derivative of x^3-9x+3

\frac{d}{d x} (x^{3} - 9 x + 3)