Evaluate $\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$

Question

I am supposed to evaluate $$\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$$ I started using partial fractions $$3x^5+13x^4+32x^3+8x^2-40x-75=x\left(x^2+3x+5^2\right)^2A+\left(x^2+3x+5^2\right)^2B+x\left(x^2+3x+5^2\right)\left(Cx+D\right)+x^2\left(Ex+F\right)$$ I managed to get to $$\:\int \:\left(\frac{2}{x}-\frac{3}{x^2}+\frac{x+1}{x^2+3x+5}+\frac{4x}{\left(x^2+3x+5\right)^2}\right)\:dx$$ Am i on the right track? is there an easier way to simplify the original integral?

Answer 1

As per the WA result from the comment section by @Alexey Burdin , your partial fraction work is found to be correct so I shall continue the work from where you got stuck. $$2\ln x +\frac{3}{x}+\underbrace{\int\frac{x+1}{x^2+3x+5}}_{I_1}+\underbrace{\int\frac{4x}{(x^2+3x+5)^2}}_{I_2}$$ Since $$I_1 =2^{-1}\int\left(\frac{2x+3}{x^2+3x+5}-\frac{1}{(\left(x+\frac{3}{2}\right)^2+\frac{11}{4}}\right)\\=\frac{1}{2}\ln\left(x^2+3x+5\right)-\frac{1}{\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)$$ And $$I_2=\int\frac{4x}{(x^2+3x+5)^2}=2\int\left(\frac{2x+3}{(x^2+3x+5)^2}-\frac{3}{(x^2+3x+5)^2}\right)\\=-\frac{2}{x^2+3x+5}-\underbrace{\int\frac{6}{(x^2+3x+5)^2}}_{I_3}$$ Notice that $$(x^2+3x+5)^2= \frac{(2x+3)^2+11}{16}$$ To evaluate the integral $I_3$ recall the reduction formula of $$\int\frac{1}{(pu^2+q)^n}=\frac{2n-3}{2q(n-1)}\int\frac{1}{(pu^2+q)^{n-1}}+\frac{u}{2q(n-1)(pu^2+q)^{n-1}}$$ with our case $p=1,q=11, u=2x+3$ and $n=2$ and hence we have $$I_3=\int\frac{-96}{((2x+3)^2+11)^2}=\frac{-96}{2}\left(\frac{1}{22}\int\frac{1}{(2x+3)^2+11}+\frac{(2x+3)}{22((2x+3)^2+11)}\right)=-\frac{24}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt {11}}\right)-\frac{24(2x+3)}{11((2x+3)^2+11)}$$ and we obtained $I_2$ as $$ -\frac{24}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{24(2x+3)}{11{(2x+3)^2+11)}}-\frac{2}{x^2+3x+5}$$ and hence we have $$2\ln x+\frac{3}{x}+\frac{1}{2}\ln(x^2+3x+5)-\frac{1}{\sqrt{11}}\tan^{-}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{2}{x^2+3x+5}-\frac{24}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{24(2x+3)}{11(2x+3)^2+11)}$$ on further simplification we yield the indefinite integral as $$\ln x^2+\frac{3}{x}+\frac{1}{2}\ln(x^2+3x+5)-\frac{35}{11\sqrt{11}}\tan^{-1}\left(\frac{2x+3}{\sqrt{11}}\right)-\frac{4(3x+10)}{11(x^2+3x+5)}$$

Answer 2

An alternative method to treat multiple roots is the following (see Application to symbolic integration): $$ \frac{3 x^5 + 13 x^4 + 32 x^3 + 8 x^2 - 40 x - 75}{x^2(x^2+3x+5)^2}=\frac{a}{x}+\frac{bx+c}{x^2+3x+5}+\frac{d}{dx}\left[\frac{e}{x}+\frac{fx+g}{x^2+3x+5}\right] $$ from which we get the linear system $$ \left\{ \begin{align} &a+b=3,\\ &6 a+3 b+c-e-f=13,\\ &19 a+5 b+3 c-6 e-2 g=32,\\ &30 a+5 c-19 e+5 f-3 g=8,\\ &25 a-30 e=-40,\\ -&25 e=-75 \end{align} \right. $$

Finally, we obtain $$ \int\left(\frac{2}{x}+\frac{x-\frac{1}{11}}{x^2+3 x+5}\right)dx+\frac{3}{x}+\frac{-\frac{12}{11}x-\frac{40}{11}}{x^2+3 x+5} $$ The subsequent integration is pretty easy.