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21

Part of 2005 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 21

Source: From Entrance Examination for Japanese University 2004

5/21/2005
[1] Tokyo Univ. of Science: ∫ln⁡x(x+1)2dx\int \frac{\ln x}{(x+1)^2}dx [2] Saitama Univ.: ∫53sin⁡x+4cos⁡xdx\int \frac{5}{3\sin x+4\cos x}dx [3] Yokohama City Univ.: ∫131x2+1dx\int_1^{\sqrt{3}} \frac{1}{\sqrt{x^2+1}}dx [4] Daido Institute of Technology: ∫0π2sin⁡3xsin⁡x+cos⁡xdx\int_0^{\frac{\pi}{2}} \frac{\sin ^ 3 x}{\sin x +\cos x}dx [5] Gunma Univ.: ∫03π4{(1+x)sin⁡x+(1−x)cos⁡x}dx\int_0^{\frac{3\pi}{4}} \{(1+x)\sin x+(1-x)\cos x\}dx
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