integral from x to xe^x of tln(t)

解

\int_{x}^{x e^{x}} t ln (t) d t

\frac{x ^{2} e ^{2 x} ln ( x ) + x ^{3} e ^{2 x} - x ^{2} ln ( x )}{2} + \frac{- x ^{2} e ^{2 x} + x ^{2}}{4}

解答ステップ

\int_{x}^{x e^{x}} t ln (t) d t

部分積分を適用する

= [\frac{1}{2} t^{2} ln (t) - \int \frac{t}{2} d t]_{x}^{x e^{x}}

\int \frac{t}{2} d t = \frac{t ^{2}}{4}

= [\frac{1}{2} t^{2} ln (t) - \frac{t ^{2}}{4}]_{x}^{x e^{x}}

限度を計算する:

\frac{1}{2} x^{2} e^{2 x} ln (x) + \frac{1}{2} x^{3} e^{2 x} - \frac{x ^{2} e ^{2 x}}{4} - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}

= \frac{1}{2} x^{2} e^{2 x} ln (x) + \frac{1}{2} x^{3} e^{2 x} - \frac{x ^{2} e ^{2 x}}{4} - \frac{1}{2} x^{2} ln (x) + \frac{x ^{2}}{4}

簡素化

= \frac{x ^{2} e ^{2 x} ln ( x ) + x ^{3} e ^{2 x} - x ^{2} ln ( x )}{2} + \frac{- x ^{2} e ^{2 x} + x ^{2}}{4}