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

解答

\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}

求解步骤

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

使用分布积分法

= [\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}}

带入上下限计算后得:

\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}

化简

= \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}

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