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Evaluate: `int(x+1)/(sqrt(4+5x-x^2)) dx` |
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Answer» Let `(x+1)=A*(d)/(dx)(4+5x-x^(2))+B. "Then", (x+1)=A(5-2x)+B`. Comparing the coefficients of like powers of x , we get `(-2A=1and5A+B=1)rArr(A=-(1)/(2)andB=(7)/(2))`. `therefore(x+1)=-(1)/(2)(5-2x)+(7)/(2)`. `thereforeint((x+1))/(sqrt(4+5x-x^(2)))dx=int(-(1)/(2)(5-2x)+(7)/(2))/(sqrt(4+5x-x^(2)))dx` `=-(1)/(2)int((5-2x))/(sqrt(4+5x-x^(2)))dx+(7)/(2)int(1)/(sqrt(4+5x-x^(2)))dx` `=-(1)/(2)int(1)/(sqrt(t))dt+(7)/(2)*int(1)/(sqrt(4-(x^(2)-5x)))dx " where "t=4+5x-x^(2)` `=-(1)/(2)*2sqrt(t)+(7)/(2)*int(dx)/(sqrt(4-(x^(2)-5x+(25)/(4))+(25)/(4)))` `=-sqrt(t)+(7)/(2)*int(dx)/(sqrt(((sqrt41)/(2))^(2)-(x-(5)/(2))^(2)))` `=-sqrt(4+5x-x^(2))+(7)/(2)*"sin"^(-1)((x-(5)/(2)))/(((sqrt(41))/(2)))+C` `=-sqrt(4+5x-x^(2))+(7)/(2)sin^(-1)((2x-5)/(sqrt(41)))+C`. |
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