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人気のある微分積分 >

integral of (ln(1+sqrt(x)))/(1+sqrt(x))

解

\int \frac{ln ( 1 + x )}{1 + x} d x

解

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

解答ステップ

\int \frac{ln ( 1 + x )}{1 + x} d x

u置換積分法を適用する

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

定数を除く:

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

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

拡張

\frac{( u - 1 ) ln ( u )}{u} : ln (u) - \frac{ln ( u )}{u}

= 2 \cdot \int ln (u) - \frac{ln ( u )}{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

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

\int ln (u) d u = u ln (u) - u

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

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

代用を戻す

u = 1 + x

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of 4e^xsec(e^x+5)

\int 4 e^{x} sec (e^{x} + 5) d x

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\int \frac{sech ^{2} ( ln ( x ) )}{x} d x

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\int 8 x^{4} + 4 x^{3} - 6 x^{2} - 4 x + 5 d x

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\int + 50 x d x

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\int \frac{7}{x ^{2} 9 a ^{2} - x ^{2}} d x