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人気のある微分積分 >

integral of xln(2x+1)

解

\int x ln (2 x + 1) d x

解

\frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3) + C

解答ステップ

\int x ln (2 x + 1) d x

u置換積分法を適用する

= \int \frac{ln ( u ) ( u - 1 )}{4} 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 ln (u) (u - 1) d u

部分積分を適用する

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

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

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

代用を戻す

u = 2 x + 1

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

簡素化

\frac{1}{4} (ln (2 x + 1) (\frac{( 2 x + 1 ) ^{2}}{2} - (2 x + 1)) - (\frac{( 2 x + 1 ) ^{2}}{4} - (2 x + 1))) : \frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3)

= \frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3)

定数を解答に追加する

= \frac{1}{16} (8 x^{2} ln (2 x + 1) - 4 x^{2} + 4 x - 2 ln (2 x + 1) + 3) + C

グラフ

人気の例

integral of (x^3+1)/(x+1)

\int \frac{x ^{3} + 1}{x + 1} d x

limit as x approaches 3-of 3/(x-3)

x \to 3 - lim (\frac{3}{x - 3})

(d^2)/(dx^2)(3/(4t^6))

\frac{d ^{2}}{d x ^{2}} (\frac{3}{4 t ^{6}})

tangent f(x)=((x-5))/((x-6),(7,2))

t an g e n t f (x) = \frac{( x - 5 )}{( x - 6 ) , ( 7 , 2 )}

derivative of 12x^3-12x^2-24x

\frac{d}{d x} (12 x^{3} - 12 x^{2} - 24 x)