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人気のある微分積分 >

integral of ln(x^2+1)x

解

\int ln (x^{2} + 1) x d x

解

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

解答ステップ

\int ln (x^{2} + 1) x d x

u置換積分法を適用する

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

定数を除く:

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

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

部分積分を適用する

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

\int 1 d u = u

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

代用を戻す

u = x^{2} + 1

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

拡張

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

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

定数を解答に追加する

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

グラフ

人気の例

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