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Question Number 121886 by oustmuchiya@gmail.com last updated on 12/Nov/20

Commented by TANMAY PANACEA last updated on 12/Nov/20

$$plzpurchaseagoodtextbookandstudythe$$$$basic$$

Answered by mathmax by abdo last updated on 12/Nov/20

$$1){\int }_1^3{({3\mathrm{x}}^3+3)}^4\mathrm{dx}=3^4{\int }_1^3{({\mathrm{x}}^3+1)}^4\mathrm{dx}$$$$=3^4{\int }_1^3\sum _{\mathrm{k}=0}^4{\mathrm{C}}_4^{\mathrm{k}}{\left({\mathrm{x}}^3\right)}^{\mathrm{k}}\mathrm{dx}=3^{\mathrm{k}}\sum _{\mathrm{k}=0}^4{\mathrm{C}}_4^{\mathrm{k}}{\int }_1^3{\mathrm{x}}^{3\mathrm{k}}\mathrm{dx}$$$$=3^{\mathrm{k}}\sum _{\mathrm{k}=0}^4{\mathrm{C}}_4^{\mathrm{k}}{\left[\frac{{\mathrm{x}}^{3\mathrm{k}+1}}{3\mathrm{k}+1}\right]}_1^3=3^{\mathrm{k}}\sum _{\mathrm{k}=0}^4\frac{{\mathrm{C}}_4^{\mathrm{k}}}{3\mathrm{k}+1}(3^{3\mathrm{k}+1}-1)$$

Answered by mathmax by abdo last updated on 12/Nov/20

$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{{2\mathrm{x}}^2-3\mathrm{x}-2}$$$${2\mathrm{x}}^2-3\mathrm{x}-2=0\to \mathrm{\Delta }=9-4(-4)=25\Rightarrow {\mathrm{x}}_1=\frac{3+5}{4}=2$$$${\mathrm{x}}_2=\frac{3-5}{4}=-\frac{1}{2}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{2\mathrm{x}+1}{2(\mathrm{x}-2)(\mathrm{x}+\frac{1}{2})}=\frac{2\mathrm{x}+1}{(2\mathrm{x}+1)(\mathrm{x}-2)}=\frac{1}{\mathrm{x}-2}$$$$\Rightarrow {\int }_3^4\frac{2\mathrm{x}+1}{{2\mathrm{x}}^2-3\mathrm{x}-2}\mathrm{dx}={\int }_3^4\frac{\mathrm{dx}}{\mathrm{x}-2}={[\ln \mid \mathrm{x}-2\mid ]}_3^4=\ln \left(2\right)$$$$$$

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