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Question Number 117945 by bemath last updated on 14/Oct/20

$$\mathrm{f}\left(\mathrm{x}\right)=\int \frac{{5\mathrm{x}}^8+{7\mathrm{x}}^6}{{({2\mathrm{x}}^7+{\mathrm{x}}^2+1)}^2}\mathrm{dx}\mathrm{and}$$$$\mathrm{f}\left(0\right)=0,\mathrm{then}\mathrm{f}\left(1\right)=\mathrm{\_}$$

Commented by bemath last updated on 14/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{all}\mathrm{sir}$$

Answered by john santu last updated on 14/Oct/20

$$f\left(x\right)=\int \frac{5x^8+7x^6}{x^{14}{(2+x^{-5}+x^{-7})}^2}dx$$$$f\left(x\right)=\int \frac{5x^{-6}+7x^{-8}}{{(2+x^{-5}+x^{-7})}^2}dx$$$$letting\varphi =2+x^{-5}+x^{-7}then$$$$d\varphi =-(5x^{-6}+7x^{-8})dx$$$$f\left(x\right)=\int \frac{-d\varphi }{{\varphi }^2}=-\int {\varphi }^{-2}d\varphi =\frac{1}{\varphi }+c$$$$f\left(x\right)=\frac{1}{2+x^{-5}+x^{-7}}+c$$$$f\left(x\right)=\frac{x^7}{2x^7+x^2+1}+c$$$$sincef\left(0\right)=0wegetc=0$$$$thusf\left(1\right)=\frac{1}{4}$$

Answered by Dwaipayan Shikari last updated on 14/Oct/20

$$\int \frac{5x^8+7x^6}{{(2x^7+x^2+1)}^2}dx$$$$=\int \frac{\frac{5}{x^6}+\frac{7}{x^8}}{{(2+\frac{1}{x^5}+\frac{1}{x^7})}^2}dx$$$$=-\int \frac{dt}{t^2}=\frac{1}{t}+C=\frac{1}{(2+\frac{1}{x^5}+\frac{1}{x^7})}$$$$f\left(x\right)=\frac{x^7}{2x^7+x^2+1}+\mathrm{C}\Rightarrow f\left(0\right)=0+C=0$$$$f\left(1\right)=\frac{1}{2.1+1+1}=\frac{1}{4}$$

Answered by 1549442205PVT last updated on 14/Oct/20

$$\mathrm{Put}\mathrm{x}=\frac{1}{\mathrm{t}}\Rightarrow \mathrm{dx}=-\frac{\mathrm{dt}}{{\mathrm{t}}^2}$$$$\mathrm{f}\left(\mathrm{x}\right)=-\int \frac{\frac{5}{{\mathrm{t}}^8}+\frac{7}{{\mathrm{t}}^6}}{{(\frac{2}{{\mathrm{t}}^7}+\frac{1}{{\mathrm{t}}^2}+1)}^2}.\frac{\mathrm{dt}}{{\mathrm{t}}^2}$$$$=-\int \frac{{\mathrm{t}}^4(5+{7\mathrm{t}}^2)\mathrm{dt}}{{({\mathrm{t}}^7+{\mathrm{t}}^5+2)}^2}=-\int \frac{{5\mathrm{t}}^4+{7\mathrm{t}}^6}{{({\mathrm{t}}^7+{\mathrm{t}}^5+2)}^2}\mathrm{dt}$$$$\mathrm{Put}{\mathrm{t}}^7+{\mathrm{t}}^5+2=\mathrm{u}\Rightarrow \mathrm{du}=({7\mathrm{t}}^6+{5\mathrm{t}}^4)\mathrm{dt}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\left(\mathrm{u}\right)\right)=-\int \frac{\mathrm{du}}{{\mathrm{u}}^2}=\frac{1}{\mathrm{u}}=\frac{1}{{\mathrm{t}}^7+{\mathrm{t}}^5+2}+\mathrm{C}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\frac{1}{{\mathrm{x}}^7}+\frac{1}{{\mathrm{x}}^5}+2}=\frac{{\mathrm{x}}^7}{{2\mathrm{x}}^7+{\mathrm{x}}^2+1}+\mathrm{C}$$$$\mathrm{f}\left(0\right)=0\Rightarrow \mathrm{C}=0.\mathrm{Hence}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^7}{{2\mathrm{x}}^7+{\mathrm{x}}^2+1}\Rightarrow \mathrm{f}\left(1\right)=\frac{1}{4}$$

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