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Question Number 108761 by bemath last updated on 19/Aug/20

$$\frac{⋮\frac{\mathcal{B}e}{\mathcal{M}ath}⋮}{★}$$$$If{\int }_{-1}^a\frac{x+1}{{(x+2)}^4}=\frac{10}{81},thenthevalueof$$$$a-2is\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Answered by ajfour last updated on 19/Aug/20

$$letI={\int }_{-1}^a\frac{x+1}{{(x+2)}^4}dx$$$$nowsayx+2=t$$$$\Rightarrow I={\int }_1^{a+2}(\frac{1}{t^3}-\frac{1}{t^4})dt=\frac{10}{81}$$$$\Rightarrow (\frac{1}{3t^3}-\frac{1}{2t^2}){\mid }_1^{a+2}=\frac{10}{81}$$$$\Rightarrow \frac{1}{3{(a+2)}^3}-\frac{1}{2{(a+2)}^2}=\frac{10}{81}+\frac{1}{6}$$$$say\frac{1}{a+2}=b$$$$\Rightarrow b^3-\frac{3}{2}{b}^2-\frac{10}{27}-\frac{1}{2}=0$$$$\Rightarrow b^3-\frac{3}{2}{b}^2-\frac{47}{54}=0$$$$or{(a+2)}^3+\frac{3}{2}\times \frac{54}{47}(a+2)-\frac{54}{47}=0$$$$saya+2=3t$$$$\Rightarrow 47t^3+9t-2=0$$$$....$$

Answered by bobhans last updated on 19/Aug/20

Commented by bemath last updated on 19/Aug/20

$$cooll$$

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