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Question Number 11429 by @ANTARES_VY last updated on 25/Mar/17

$$\underset{0}{\stackrel{3}{\int }}\frac{2\mathbf{x}+3}{2\mathbf{x}+1}\mathbf{dx}=\mathit{\alpha }+\mathbf{ln}7.$$$$\mathit{\alpha }=?$$$$\mathbf{please}......$$

Answered by FilupS last updated on 26/Mar/17

$$\frac{2x+3}{2x+1}=\frac{2x+1+2}{2x+1}=1+\frac{2}{2x+1}$$$${\int }_0^3\frac{2x+3}{2x+1}dx={\int }_0^{3}1+\frac{2}{2x+1}dx$$$$={\int }_0^{3}1dx+{\int }_0^3\frac{2}{2x+1}dx$$$$={\left[x\right]}_0^3+{\left[\ln (2x+1)\right]}_0^3$$$$=(3-0)+(\ln \left(7\right)-\ln \left(1\right))$$$$=3+\ln \left(7\right)$$$$$$$$\therefore \alpha =3$$$$$$$$\mathrm{red}\mathrm{text}\mathrm{is}\mathrm{corrected}$$

Commented by @ANTARES_VY last updated on 26/Mar/17

$$\mathit{\alpha }=2?...$$

Commented by FilupS last updated on 26/Mar/17

$$\mathrm{yes}$$

Commented by @ANTARES_VY last updated on 26/Mar/17

$$\mathbf{such}\mathbf{a}\mathbf{wrong}\mathbf{answer}....$$$$\mathbf{no}\mathit{\alpha }=2?????$$

Commented by FilupS last updated on 26/Mar/17

$${\int }_0^3\frac{2x+3}{2x+1}dx=\alpha +\ln \left(7\right)$$$$\mathrm{as}\mathrm{per}\mathrm{my}\mathrm{working}:$$$${\int }_0^3\frac{2x+3}{2x+1}dx=2+\ln \left(7\right)$$$$\therefore \alpha +\ln \left(7\right)=2+\ln \left(7\right)$$$$\alpha =2$$

Commented by ajfour last updated on 26/Mar/17

$$\mathrm{come}\mathrm{on}3+\ln 7\mathrm{it}\mathrm{is}.$$$${\int }_0^3\frac{2\mathrm{x}+3}{2\mathrm{x}+1}\mathrm{dx}={\int }_0^3\mathrm{dx}+{\int }_0^3\frac{2\mathrm{dx}}{2\mathrm{x}+1}$$

Commented by FilupS last updated on 26/Mar/17

$$\mathrm{I}\mathrm{see}\mathrm{my}\mathrm{mistake}.$$$$\mathrm{I}\mathrm{made}\mathrm{a}\mathrm{bad}\mathrm{typo}.\mathrm{Sorry}$$

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