Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 4

Use integration by parts with $u=\ln (1+x^2)$ and $dv=(1+3x^2)dx$, giving $v=x+x^3=x(1+x^2)$. The remaining integrand simplifies because $du=2xdx/(1+x^2)$:The real formula is valid for every real number $x$.

Put $u=\ln x$, so $du=dx/x$. The integral becomes $\int du/u$, givingFor the real logarithm, choose an interval in $0<x<1$ or $x>1$; the original integrand is singular at one and undefined for nonpositive $x$.

Put $u=\cos x$, $du=-\sin xdx$, and use ${\sin }^{2}x=1-u^2$. ThenThereforeThis holds on intervals where $\cos x\ne 0$.