A) \[\left( -\frac{8}{3},\,-\frac{2}{3} \right)\]
B) \[\left( -\frac{8}{3},\,\frac{2}{3} \right)\]
C) \[\left( \frac{8}{3},\,\frac{2}{3} \right)\]
D) \[\left( \frac{8}{3},\,-\frac{2}{3} \right)\]
Correct Answer: B
Solution :
\[f\left( \frac{3x-4}{3x+4} \right)\,=x+2,\,\,x\ne -\frac{4}{3}\] Let \[\frac{3x-4}{3x+4}\,=t\] \[3x-4=3tx+4t\] \[x=\frac{4t+4}{3-3t}+2\] \[f(t)\,=\frac{10-2t}{3-3t}\] \[f(x)=\frac{2x-10}{3x-3}\] \[\int_{{}}^{{}}{f(x)\,dx=\int_{{}}^{{}}{\frac{2x-10}{3x-3}dx}}\] \[=\int_{{}}^{{}}{\frac{2x}{3x-3}\,dx-10\,\int_{{}}^{{}}{\frac{dx}{3x-3}}}\] \[=\,\frac{2}{3}\,\int_{{}}^{{}}{\frac{x-1}{x-1}\,dx+\frac{2}{3}\,\int_{{}}^{{}}{\frac{dx}{x-1}\,-\frac{10}{3}\,\int_{{}}^{{}}{\frac{dx}{x-1}\,=\frac{2x}{3}-\frac{8}{3}\ln \,(x-1)+C}}}\]
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