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人気のある微分積分 >

integral of xln(2+x^2)

解

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

解

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

解答ステップ

\int x ln (2 + x^{2}) 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 = 2 + x^{2}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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