integral from 0 to a of arcsinh(t)

解答

\int_{0}^{a} arcsinh (t) d t

a arcsinh (a) - a^{2} + 1 + 1

求解步骤

\int_{0}^{a} arcsinh (t) d t

使用分布积分法

= [t arcsinh (t) - \int \frac{t}{t ^{2} + 1} d t]_{0}^{a}

\int \frac{t}{t ^{2} + 1} d t = t^{2} + 1

= [t arcsinh (t) - t^{2} + 1]_{0}^{a}

带入上下限计算后得:

a arcsinh (a) - a^{2} + 1 + 1

= a arcsinh (a) - a^{2} + 1 + 1

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