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Question Number 37425 by MJS last updated on 13/Jun/18

$$\int \frac{x^5-3x^4-23x^3+51x^2+94x-120}{8x^3\sqrt{42+x-x^2}}dx=?$$

Answered by ajfour last updated on 13/Jun/18

$$d(42+x-x^2)=(1-2x)dx$$$$I=\frac{1}{8}\int \frac{x^2-3x-23}{\sqrt{42+x-x^2}}dx$$$$+\frac{1}{8}\int \frac{51x^2+94x-120}{x^3\sqrt{42+x-x^2}}dx$$$$8I=I_1+I_2$$$$forsecondintegralletx=\frac{1}{t}$$$$I_2=\int \frac{51x^2+94x-120}{x^3\sqrt{42+x-x^2}}dx$$$$=-\int \frac{t^3(\frac{51}{t^2}+\frac{94}{t}-120)dt}{t^2\sqrt{42+\frac{1}{t}-\frac{1}{t^2}}}$$$$=\int \frac{120t^2-94t-51}{\sqrt{42t^2+t-1}}dt$$$$120t^2-94t-51=\mathit{C}(42t^2+t-1)$$$$+\mathit{D}(84t+1)$$$$similarlyfor{I}_1:$$$$x^2-3x-23=\mathit{A}(42+x-x^2)$$$$+\mathit{B}(1-2x)$$$$8I=\mathit{A}\int \sqrt{42+x-x^2}dx$$$$+\mathit{B}\int \frac{(1-2x)dx}{\sqrt{42+x-x^2}}$$$$+\mathit{C}\int \sqrt{42t^2+t-1}dt$$$$+\mathit{D}\int \frac{(84t+1)dt}{\sqrt{42t^2+t-1}}.$$$$$$$$.......$$

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