Question Number 22946 by ANTARES_VY last updated on 24/Oct/17
$$\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\frac{\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{2}}\boldsymbol{\mathrm{dx}}\:\:\:?\boldsymbol{\mathrm{solve}} \\ $$
Answered by ajfour last updated on 24/Oct/17
$$=\mathrm{ln}\:\mathrm{19}\: \\ $$
Answered by Joel577 last updated on 24/Oct/17
![Let u = 2x^2 + x − 2 → du = 4x + 1 dx x = 1 → u = 1 x = 3 → u = 19 I = ∫_1 ^(19) ((4x + 1)/u) . (du/(4x + 1)) = [ln u]_1 ^(19) = ln 19 − ln 1 = ln 19](Q22970.png)
$$\mathrm{Let}\:{u}\:=\:\mathrm{2}{x}^{\mathrm{2}} \:+\:{x}\:−\:\mathrm{2}\:\:\rightarrow\:\:{du}\:=\:\mathrm{4}{x}\:+\:\mathrm{1}\:{dx} \\ $$$${x}\:=\:\mathrm{1}\:\:\rightarrow\:\:{u}\:=\:\mathrm{1} \\ $$$${x}\:=\:\mathrm{3}\:\:\rightarrow\:\:{u}\:=\:\mathrm{19} \\ $$$${I}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{19}} {\int}}\:\:\frac{\mathrm{4}{x}\:+\:\mathrm{1}}{{u}}\:.\:\frac{{du}}{\mathrm{4}{x}\:+\:\mathrm{1}} \\ $$$$\:\:\:\:=\:\left[\mathrm{ln}\:{u}\right]_{\mathrm{1}} ^{\mathrm{19}} \:=\:\mathrm{ln}\:\mathrm{19}\:−\:\mathrm{ln}\:\mathrm{1}\:=\:\mathrm{ln}\:\mathrm{19} \\ $$