Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 13

The hyperbolic identity ${\cosh }^{2}x-{\sinh }^{2}x=1$ and $\cosh x+\sinh x=e^x$ reduce the integrand to $1+e^x$. Therefore

Let $D=3x^2+2x+2$. Splitting the numerator into a derivative term and a constant gives $x-1=D^{\prime }/6-4/3$. ThusComplete the square as $D=((3x+1{)}^2+5)/3$ and substitute $u=3x+1$. The remaining integral is $\int du/(u^2+5)=\arctan (u/\sqrt{5})/\sqrt{5}$. HenceThe quadratic denominator is positive for all real $x$, so no logarithmic absolute value or real-domain exclusion is needed.

The argument $3x^3+2x$ is odd, so its squared sine is even. Replacing $x$ by $-x$ interchanges the numerator and denominator inside the logarithm, changing that factor's sign. Thus the full integrand is an odd function. It is continuous on the symmetric interval because $|x|\le 1/\pi <1$. Therefore

The fundamental theorem of calculus states that integrating a continuous function from a fixed lower endpoint produces an antiderivative. More precisely, if $f$ is continuous at the interior point $x$, thenIndeed, the difference quotient is the average of $f$ over the short interval from $x$ to $x+h$, which tends to $f(x)$ by continuity. The accompanying evaluation rule is ${\int }_a^{b}f=F(b)-F(a)$ whenever $F^{\prime }=f$ on the interval.

The sum is finite, so differentiation passes through it without a convergence issue. Each lower limit $n$ is constant with respect to $x$, and the fundamental theorem of calculus gives the same upper-end integrand. Hence the derivative isThe binomial theorem at $1+1$ gives the sum $2^N$. Therefore