求∫x-3/x²-2x+2 dx,∫x³/√(4-x²)dx
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![求∫x-3/x²-2x+2 dx,∫x³/√(4-x²)dx](/uploads/image/z/3819208-40-8.jpg?t=%E6%B1%82%E2%88%ABx-3%2Fx%26%23178%3B-2x%2B2+dx%2C%E2%88%ABx%26%23179%3B%2F%E2%88%9A%EF%BC%884-x%26%23178%3B%EF%BC%89dx)
求∫x-3/x²-2x+2 dx,∫x³/√(4-x²)dx
求∫x-3/x²-2x+2 dx,∫x³/√(4-x²)dx
求∫x-3/x²-2x+2 dx,∫x³/√(4-x²)dx
1,=∫(x-1-2)/((x-1)^2+1)dx
=1/2∫1/((x-1)^2+1)d((x-1)^2+1)-2∫1/((x-1)^2+1)d(x-1)
=1/2ln(x^2-2x+2)-2arctg(x-1)+c
2,=1/2∫((4-x^2)-4)/√(4-x²)d(4-x^2)
=1/3(4-x^2)^3/2-4(4-x^2)^1/2+c