| > (1)∫[0→π/4]sin^3θcos^2θdθ =∫[0→π/4]sinθcos^2θ(1-cos^2θ)dθ > (2)∫[0→√3]{1/((x^(2)+1)^(3/2))}dx x=tanθと置換 > (3)∫[0→1]xlog(2x+1)dx=[(x^2/2)log(2x+1)][0→1]-∫[0→1](x^2/2)2/(2x+1)dx(部分積分) =[(x^2/2)log(2x+1)][0→1]-∫[0→1]{x(2x+1)/2-(2x+1)/4+1/4}/(2x+1)dx =[(x^2/2)log(2x+1)][0→1]-∫[0→1]{x/2-1/4+(1/4)/(2x+1)}dx=…
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