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Question Number 22946 by ANTARES_VY last updated on 24/Oct/17

$$\underset{1}{\stackrel{3}{\int }}\frac{4\mathbf{x}+1}{2{\mathbf{x}}^2+\mathbf{x}-2}\mathbf{dx}?\mathbf{solve}$$

Answered by ajfour last updated on 24/Oct/17

$$=\ln 19$$

Answered by Joel577 last updated on 24/Oct/17

$$\mathrm{Let}u=2x^2+x-2\to du=4x+1dx$$$$x=1\to u=1$$$$x=3\to u=19$$$$I=\underset{1}{\stackrel{19}{\int }}\frac{4x+1}{u}.\frac{du}{4x+1}$$$$={\left[\ln u\right]}_1^{19}=\ln 19-\ln 1=\ln 19$$

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