Question

Integrate x+1/(x^2+1)x dx

\(\int\frac{x+1}{(x^2+1)x}dx\)

Answer 1

\(\int\frac{(x+1)dx}{x^2+1}=\frac12\int\frac{2x}{x^2+1}dx\) + \(\int\frac1{x^2+1}dx\) 

 = \(\frac12ln|x^2+1|+tan^{-1}x+c\)

Answer 2

\(\int\frac{x+1}{(x^2+1)x}dx\) 

 = \(\int\frac{x}{(x^2+1)x}dx+\int\frac1{x(x^2+1)}dx\)

\(=\int\frac1{1+x^2}dx+\int\cfrac{1}{x^3(1+\frac1{x^2})}dx\) 

\(=tan^{-1}x+\int\cfrac{-\frac{dt}2}{t}\) 

(\(\int\frac{dx}{1+x^2}=tan^{-1}x\) and by taking 1 + \(\frac1{x^2}=t\) ⇒ \(\frac{-2}{x^3}dx=dt\))

= tan^-1x - \(\frac12ln(1+\frac1{x^2})+c\) (\(\because t=1+\frac1{x^2}\)) 

Hence, \(\int\frac{x+1}{(x^2+1)x}dx=tan^{-1}x-\frac12ln(1+\frac1{x^2})+c\)