Question

integration of (x+1)dx/x(1+xe^x)^2

Answer 1

hope it help you dear.....

Answer 2

The integration of the given function is,

$\int\limits \,\frac{x+1}{x(1+xe^x)^2} dx = log(\frac{xe^x}{1 + xe^x} ) + \frac{1}{1+xe^x}+C$

• Now, we have

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

• Let

                $1 + xe^x = t\\xe^x = t-1$        - (2)

         differentiating with respect to 'x', we get

               $(e^x + xe^x ) dx = dt \\e^x(1 +x) dx = dt\\(1+x)dx=\frac{dt}{e^x}$

• Now,  We can write (1) as

                 $\int\limits \,\frac{1}{xe^xt^2} dt$

       now, from (2) :

           $\int\limits \,\frac{1}{(t-1)t^2} dt$      -(3)

• By using integration by parts ,   $\frac{1}{(t-1)t^2}$ can be written as,

          $\frac{1}{(t-1)t^2} = \frac{1}{t-1} - \frac{t+1}{t^2}$

          therefore, (3) becomes,

          $\int\limits \, \frac{1}{(t-1)t^2}dt = \int\limits \, \frac{1}{t-1}dt - \int\limits \, \frac{t+1}{t^2}dt$

             $= log(t-1) - logt + \frac{1}{t} + C$

             $= log(\frac{t-1}{t} ) + \frac{1}{t}+C$

• Now substituting the value of 't' from (2), we get

             $= log(\frac{xe^x}{1 + xe^x} ) + \frac{1}{1+xe^x}+C$

        therefore we get the integration of the given function as,

           $\int\limits \,\frac{x+1}{x(1+xe^x)^2} dx = log(\frac{xe^x}{1 + xe^x} ) + \frac{1}{1+xe^x}+C$