Question

$\int\limits^3_1 dx/x²(x+1) ,Evaluate it.$

Answer 1

Answer:

$log\frac{2}{3}-\frac{4}{3}$

Step-by-step explanation:

$\int\limits^3_1 {\frac{1}{x^{2}(x+1)}} \, dx$

$\textrm{using partial fractions}$

$\frac{A}{x+1}+\frac{Bx+C}{x^{2}}\textrm{where A,B,C are constant}$

$\textrm{Taking L.C.M we get}$

$\frac{(A+B)x^{2}+(B+C)x+C}{x^{2}x(x+1)}$

$\textrm{on solving we get A=1,B=-1,C=1}$

$\int\limits^3_1 {\frac{1}{(x+1)}} \, dx+\int\limits^3_1 {\frac{-x+1}{x^{2}}} \, dx$

$\int\limits^3_1 {\frac{1}{(x+1)}} \, dx-\int\limits^3_1 {\frac{x}{x^{2}}} \, dx+\int\limits^3_1 {\frac{1}{x^{2}}} \, dx$

$[tex]\int\limits^3_1 {\frac{1}{(x+1)}} \, dx-\int\limits^3_1 {\frac{1}{x}} \, dx+\int\limits^3_1 {\frac{1}{x^{2}}} \, dx$

$\textrm{applying thr formula of integrals,we get}$

$\left \{ {{x=3} \atop {x=1}} \right.log(x+1)-log(x)-\frac{1}{x}+c$

$\textrm{on applying the upper and lower limits we get,}$

$log\frac{2}{3}-\frac{4}{3}$