integral of (e^{1/(x^2)}+2x+ln(x))/(x^3)

解

\int \frac{e ^{\frac{1}{x ^{2}}} + 2 x + ln ( x )}{x ^{3}} d x

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

解答ステップ

\int \frac{e ^{\frac{1}{x ^{2}}} + 2 x + ln ( x )}{x ^{3}} d x

拡張

\frac{e ^{\frac{1}{x ^{2}}} + 2 x + ln ( x )}{x ^{3}} : \frac{e ^{\frac{1}{x ^{2}}}}{x ^{3}} + \frac{2}{x ^{2}} + \frac{ln ( x )}{x ^{3}}

= \int \frac{e ^{\frac{1}{x ^{2}}}}{x ^{3}} + \frac{2}{x ^{2}} + \frac{ln ( x )}{x ^{3}} d x

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

= \int \frac{e ^{\frac{1}{x ^{2}}}}{x ^{3}} d x + \int \frac{2}{x ^{2}} d x + \int \frac{ln ( x )}{x ^{3}} d x

\int \frac{e ^{\frac{1}{x ^{2}}}}{x ^{3}} d x = - \frac{1}{2} e^{\frac{1}{x ^{2}}}

\int \frac{2}{x ^{2}} d x = - \frac{2}{x}

\int \frac{ln ( x )}{x ^{3}} d x = - \frac{ln ( x )}{2 x ^{2}} - \frac{1}{4 x ^{2}}

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

定数を解答に追加する

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