Evaluate the six rational antiderivatives \[ \int\frac{x}{x-1}\,dx, \quad \int\frac{7x+5}{x^2+x-2}\,dx, \quad \int\frac{x}{x^2-x-6}\,dx, \] \[ \int\frac{1+x^2}{1-x^2}\,dx, \quad \int\frac{dx}{x(x-1)^2}, \quad \int\frac{x}{(x+5)^2}\,dx. \]
Solution
Using elementary partial fractions, \[ \frac{x}{x-1}=1+\frac1{x-1}, \] so \[ \boxed{x+\ln|x-1|+C.} \] Also \[ \frac{7x+5}{(x+2)(x-1)}=\frac3{x+2}+\frac4{x-1}, \] hence \[ \boxed{3\ln|x+2|+4\ln|x-1|+C.} \] Next, \[ \frac{x}{(x-3)(x+2)}=\frac{3/5}{x-3}+\frac{2/5}{x+2}, \] so \[ \boxed{\frac35\ln|x-3|+\frac25\ln|x+2|+C.} \] Further, \[ \frac{1+x^2}{1-x^2}=-1+\frac1{x+1}-\frac1{x-1}, \] therefore \[ \boxed{-x+\ln|x+1|-\ln|x-1|+C.} \] For the repeated factor, \[ \frac1{x(x-1)^2}=\frac1x-\frac1{x-1}+\frac1{(x-1)^2}, \] which gives \[ \boxed{\ln|x|-\ln|x-1|-\frac1{x-1}+C.} \] Finally, \[ \frac{x}{(x+5)^2}=\frac1{x+5}-\frac5{(x+5)^2}, \] so \[ \boxed{\ln|x+5|+\frac5{x+5}+C.} \]