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Today's calculation of Integral 179

Source: From 2 university entrance exams 2007

February 8, 2007
calculusintegrationlogarithmscalculus computations

Problem Statement

Evaluate the following integrals. (1) Meiji University ∫1ee(log⁡x)2xdx.\int_{\frac{1}{e}}^{e}\frac{(\log x)^{2}}{x}dx. (2) Tokyo University of Science ∫017x3+23x2+21x+15(x2+1)(x+1)2dx.\int_{0}^{1}\frac{7x^{3}+23x^{2}+21x+15}{(x^{2}+1)(x+1)^{2}}dx.
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