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Question Number 59679 by Forkum Michael Choungong last updated on 13/May/19

$$Evaluate$$$$\underset{0}{\stackrel{3}{\int }}\left(\frac{x^2+3x}{x^3}\right)$$$$$$

Answered by Kunal12588 last updated on 13/May/19

$$\underset{0}{\stackrel{3}{\int }}\left(\frac{x^2+3x}{x^3}\right)dx$$$$=\underset{0}{\stackrel{3}{\int }}(\frac{x^2}{x^3}+\frac{3x}{x^3})dx$$$$={\int }_0^3\frac{1}{x}dx+3{\int }_0^3\frac{1}{x^2}dx$$$$={\left[ln\left(x\right)\right]}_0^3+3{\left[\frac{x^{-2+1}}{-2+1}\right]}_0^3$$$$=ln\left(3\right)-ln\left(0\right)-3(3^{-1}-0^{-1})$$$$so{\int }_1^3\left(\frac{x^2+3x}{x^3}\right)dxisundefined$$

Answered by meme last updated on 13/May/19

$${\int }_0^3\frac{x+3}{x^2}={\int }_0^3\frac{1}{x}+{\int }_0^3\frac{1}{x^2}={\left[ln\left(x\right)\right]}_0^3-3{\left[\frac{1}{x}\right]}_0^3$$$$impossible\left(ln0\right)$$

Answered by Joel122 last updated on 14/May/19

$$\mathrm{Let}f\left(x\right)=\frac{x^2+3x}{x^3}$$$$f\left(x\right)\mathrm{is}\mathrm{undefined}\mathrm{at}x=0,\mathrm{so}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{improper}\mathrm{integral}$$$$I=\underset{0}{\stackrel{3}{\int }}\frac{x^2+3x}{x^3}dx=\underset{a\to 0^{-}}{\lim }\left[\underset{a}{\stackrel{3}{\int }}\frac{x^2+3x}{x^3}dx\right]$$$$\left[\begin{array}{c}\underset{a}{\stackrel{3}{\int }}\frac{x^2+3x}{x^3}dx=\underset{a}{\stackrel{3}{\int }}\frac{1}{x}+\frac{3}{x^2}dx={[\ln x-\frac{3}{x}]}_a^3\\ =\ln \left(\frac{3}{a}\right)+\frac{3}{a}-1\end{array}\right]$$$$I=\underset{a\to 0^{-}}{\lim }[\ln \left(\frac{3}{a}\right)+\frac{3}{a}-1]=\mathrm{\infty }$$$$\therefore \mathrm{The}\mathrm{integral}\mathrm{is}\mathrm{divergent}$$

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