integral from 1 to x of ln(8+t^2)

解答

\int_{1}^{x} ln (8 + t^{2}) d t

x ln (8 + x^{2}) + 42 arctan (\frac{x}{2 2}) - 2 x - 2 ln (3) - 42 arctan (\frac{1}{2 2}) + 2

求解步骤

\int_{1}^{x} ln (8 + t^{2}) d t

使用分布积分法

= [t ln (8 + t^{2}) - \int \frac{2 t ^{2}}{8 + t ^{2}} d t]_{1}^{x}

\int \frac{2 t ^{2}}{8 + t ^{2}} d t = - 42 arctan (\frac{t}{2 2}) + 2 t

= [t ln (8 + t^{2}) - (- 42 arctan (\frac{t}{2 2}) + 2 t)]_{1}^{x}

化简

= [t ln (8 + t^{2}) + 42 arctan (\frac{t}{2 2}) - 2 t]_{1}^{x}

带入上下限计算后得:

x ln (8 + x^{2}) + 42 arctan (\frac{x}{2 2}) - 2 x - 2 ln (3) - 42 arctan (\frac{1}{2 2}) + 2

= x ln (8 + x^{2}) + 42 arctan (\frac{x}{2 2}) - 2 x - 2 ln (3) - 42 arctan (\frac{1}{2 2}) + 2

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