Q.Integrate the following function: $\frac{2x}{(x^2+1)(x^2+3)}$

This question offers a choice. Alternative 1 integrates a rational function by partial fractions. Alternative 2 evaluates a definite integral of an absolute-value function by splitting at the point where the expression inside changes sign.

Alternative 1 — integrate $\dfrac{2x-3}{(x^2-1)(2x+3)}$:

Factor $x^2-1=(x-1)(x+1)$, so the denominator is $(x-1)(x+1)(2x+3)$. Write the partial fraction decomposition:

$\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}$

Multiplying through:

$2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)$

Plug in convenient values of $x$:

• $x=1$: $2-3=-1=A(2)(5)=10A\Rightarrow A=-\dfrac1{10}$
• $x=-1$: $-2-3=-5=B(-2)(1)=-2B\Rightarrow B=\dfrac52$
• $x=-\dfrac32$: $-3-3=-6=C\left(-\dfrac52\right)\left(-\dfrac12\right)=C\cdot\dfrac54\Rightarrow C=-\dfrac{24}{5}$

So:

$\int\frac{2x-3}{(x^2-1)(2x+3)}dx=\int\left[\frac{-1/10}{x-1}+\frac{5/2}{x+1}+\frac{-24/5}{2x+3}\right]dx$

$=-\frac{1}{10}\ln|x-1|+\frac{5}{2}\ln|x+1|-\frac{24}{5}\cdot\frac12\ln|2x+3|+C$

$=-\frac{1}{10}\ln|x-1|+\frac{5}{2}\ln|x+1|-\frac{12}{5}\ln|2x+3|+C$

Alternative 2 — evaluate $\displaystyle\int_{-5}^{5}|x+2|\,dx$:

The expression $x+2$ changes sign at $x=-2$, which lies inside $[-5,5]$. So split the integral there: …

Substitute $t=\sin x$, then split into partial fractions.

Main part. Let $t=\sin x$, so $dt=\cos x\,dx$. The integral becomes $\displaystyle\int \dfrac{dt}{(1-t)(2-t)}$.

Partial fractions: $\dfrac{1}{(1-t)(2-t)} = \dfrac{A}{1-t}+\dfrac{B}{2-t}$. Solving, $A=1,\ B=-1$.

$\displaystyle\int\left[\dfrac{1}{1-t}-\dfrac{1}{2-t}\right]dt = -\ln|1-t| + \ln|2-t| + C = \ln\left|\dfrac{2-t}{1-t}\right|+C$

Substituting back $t=\sin x$: $\ln\left|\dfrac{2-\sin x}{1-\sin x}\right| + C$.

OR (alternative part). $\displaystyle\int_0^2(x^2+1)dx$ as a limit of a sum: with $h=\dfrac{2-0}{n}=\dfrac2n$,

$\displaystyle\int_0^2 f(x)dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(rh)$, $f(x)=x^2+1$.

$S_n = h\sum_{r=0}^{n-1}\left[(rh)^2+1\right] = h^3\sum r^2 + hn = h^3\cdot\dfrac{(n-1)n(2n-1)}{6} + hn$

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