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人気のある微分積分 >

integral of xln(x^2+1)

解

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

inverselaplace 1/(s^2(s+1.1))

in v erse l a pl a ce \frac{1}{s ^{2} ( s + 1.1 )}

integral of 1/(sqrt(2+x-x^2))

\int \frac{1}{2 + x - x ^{2}} d x

integral of xsqrt(x^2-5)

\int x x^{2} - 5 d x

limit as x approaches-1-of arccos(x)

x \to - 1 - lim (arccos (x))

derivative of 32x

\frac{d}{d x} (32 x)