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Prove the following (Exercises 34 to 39)

    Q34.    \int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}

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best_answer

L.H.S = \int_1^3\frac{dx}{x^2(x+1)}

We can write the numerator as [(x+1) -x]

\therefore \int_1^3\frac{dx}{x^2(x+1)} = \int_1^3\frac{(x+1)-x}{x^2(x+1)}dx

\\ = \int_1^3\left [ \frac{1}{x^2} - \frac{1}{x(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{(x+1)-x}{x(x+1)}dx

\\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\left [ \frac{1}{x} - \frac{1}{(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{1}{x}dx + \int_1^3\frac{1}{(x+1)}dx \\ = \left [ -\frac{1}{x} \right ]^3_1 - \left [ \log x \right ]^3_1 +\left [ \log(x+1) \right ]^3_1

\\ = \left [ -\frac{1}{3} +1 \right ] - \left [ \log 3 - \log 1 \right ] +\left [\log 4 - \log 2 \right ] \\ = \frac{2}{3} + \log \left ( \frac{4}{3.2}\right ) \\ 

= \log \left(\frac{2}{3} \right ) +\frac{2}{3}    = RHS

Hence proved.

Posted by

HARSH KANKARIA

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