Compute, on each interval where the integrand is defined, \[ \int \frac{x}{x^4-16}\,dx, \qquad \int \frac{x\arctan x}{(x^2+1)^2}\,dx. \]
Solution
1. Integrate \(x/(x^4-16)\).
Let \(u=x^2\), so \(du=2x\,dx\). Then \[ \int\frac{x}{x^4-16}\,dx =\frac12\int\frac{du}{u^2-16}. \] Since \[ \frac1{u^2-16}=\frac18\left(\frac1{u-4}-\frac1{u+4}\right), \] we get \[ \boxed{ \int\frac{x}{x^4-16}\,dx =\frac1{16}\log\left|\frac{x^2-4}{x^2+4}\right|+C.} \]
2. Integrate \(x\arctan x/(1+x^2)^2\).
Integrate by parts with \[ u=\arctan x, \qquad dv=\frac{x\,dx}{(1+x^2)^2}. \] Then \[ du=\frac{dx}{1+x^2}, \qquad v=-\frac1{2(1+x^2)}, \] so \[ I=-\frac{\arctan x}{2(1+x^2)} +\frac12\int\frac{dx}{(1+x^2)^2}. \] For the remaining integral, put \(x=\tan t\). Then \[ \int\frac{dx}{(1+x^2)^2} =\int\cos^2t\,dt =\frac t2+\frac{\sin2t}{4}+C. \] Since \(t=\arctan x\) and \(\sin2t=2x/(1+x^2)\), \[ \int\frac{dx}{(1+x^2)^2} =\frac12\arctan x+\frac{x}{2(1+x^2)}+C. \] Therefore \[ \boxed{ \int\frac{x\arctan x}{(1+x^2)^2}\,dx =-\frac{\arctan x}{2(1+x^2)} +\frac{x}{4(1+x^2)} +\frac14\arctan x+C.} \]