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人気のある微分積分 >

integral of 4(x-1)ln(x-1)

解

\int 4 (x - 1) ln (x - 1) d x

解

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

解答ステップ

\int 4 (x - 1) ln (x - 1) d x

定数を除く:

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

= 4 \cdot \int (x - 1) ln (x - 1) d x

拡張

(x - 1) ln (x - 1) : x ln (x - 1) - ln (x - 1)

= 4 \cdot \int x ln (x - 1) - ln (x - 1) d x

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

= 4 (\int x ln (x - 1) d x - \int ln (x - 1) d x)

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

\int ln (x - 1) d x = x ln (x - 1) - ln (x - 1) - x + 1

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

y^'-3/x y=(y^4)/(x^7)

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{7}}

derivative of 4sin^2(3x)

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

y^{''}+10y^'+25y=e^{-4x}(x^2+2x-4)

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{- 4 x} (x^{2} + 2 x - 4)

derivative of cos^2(e^x)

\frac{d}{d x} (cos^{2} (e^{x}))

勾配 (7,-2),(3,5)

s l o p e (7, - 2), (3, 5)