$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition

Question

One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\frac{2}{(x^2+1)^2}dx$$ and then proceed by using partial fraction decomposition on the first integral. The second integral could be dealt with by substituting $x=\tan \theta$.

Is there a way to evaluate this integral without using partial fractions, and preferably without splitting it into two integrals as I did here?

Answer 1

Let $$\frac{x^2+1}{(x+1)^2}=t\implies x=\frac{\sqrt{2 t-1}-t}{t-1}\implies dx=\frac{dt}{1-t \left(\sqrt{2 t-1}+2\right)}$$ to make $$I=\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx=\int\left(-\frac{1}{2 t^2}+\frac{1}{4 t}-\frac{3}{4 t \sqrt{2 t-1}}\right)\,dt$$ which does not seem too bad. $$I=\frac{1}{2 t}+\frac 14 \log(t)-\frac{3}{2} \tan ^{-1}\left(\sqrt{2 t-1}\right)+C$$

Answer 2

$$\operatorname*{Res}_{z=-1}\frac{z^3+3z+2}{(z^2+1)^2(z+1)}=\lim_{z\to -1}\frac{z^3+3z+2}{(z^2+1)^2}=-\frac{1}{2}$$ so we know in advance that the integrand function plus $\frac{1}{2(x+1)}$ can be written as $\frac{p(x)}{(x^2+1)^2}$:

$$ \frac{x^3+3x+2}{(x^2+1)^2(x+1)}+\frac{1}{2(x+1)} = \frac{x^3+x^2+x+5}{2(x^2+1)^2}=\frac{x}{2(x^2+1)}+\frac{x^2+5}{2(x^2+1)^2}. $$ This leads to the decomposition $$ \frac{x^3+3x+2}{(x^2+1)^2(x+1)}=-\frac{1}{2(x+1)}+\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+1)}+\frac{2}{(x^2+1)^2} $$ and also to $$\int\frac{x^3+3x+2}{(x^2+1)^2(x+1)}\,dx= -\frac{1}{2}\log(x+1)+\frac{1}{4}\log(x^2+1)+\frac{1}{2}\arctan(x)+2\int\frac{dx}{(x^2+1)^2} $$ where the last integral is immediately solved by $x\mapsto\tan\theta$.

Answer 3

$$\begin{align*} & \int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx \\ &= \int \frac{\tan^3t+3\tan t+2}{\left(\tan^2t+1\right)^2(\tan t+1)} \, \sec^2t \, dt \tag 1 \\ &= \int \frac{2\left(\cos^3t-\sin^3t\right)+3\sin t}{\sin t+\cos t} \, dt \\ &= 2 \int \frac{\cos\left(t+\frac\pi4\right) \left(1 + \sin t\cos t\right)}{\sin\left(t+\frac\pi4\right)} \, dt + \frac3{\sqrt2} \int \frac{\sin t}{\sin\left(t+\frac\pi4\right)} \, dt \tag2 \\ &= \int \frac{u^2+1}u \, du + \frac3{\sqrt2} \int \frac{\sin\left(v-\frac\pi4\right)}{\sin v} \, dv \tag3 \\ &= \int\left(u+\frac1u\right) \,du + \frac32 \int (1-\cot v) \, dv \\ &= \frac{u^2}2 + \ln|u| + \frac{3v}2 - \frac32 \ln\left|\sin v\right| + C \\ &= \frac{(\sin t+\cos t)^2}2 + \ln\left|\sin t+\cos t\right| + \frac32\left(t+\frac\pi4\right) - \frac32 \ln\left|\sin\left(t+\frac\pi4\right)\right| + C \\ &= \frac32t + \frac12\sin(2t) - \frac12 \ln\left|\sin t+\cos t\right| + C \\ &= \frac32 \tan^{-1}x + \frac x{1+x^2} - \frac12 \ln\left|\frac{x+1}{\sqrt{x^2+1}}\right| + C \end{align*}$$

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• $(1)$ : substitute $x=\tan t$
• $(2)$ : recall $\sin(a+b)=\sin a\cos b+\cos a\sin b$
• $(2)$ : substitute $u=\sin\left(t+\dfrac\pi4\right)$ and $v=t+\dfrac\pi4$. Note that

$$\sin t+\cos t=\sqrt2 \sin\left(t+\frac\pi4\right) = u \\ \implies \sin t-\cos t = \sqrt2 \cos\left(t+\frac\pi4\right) = \sqrt{2-u^2} \\ \implies \sin t=\frac{u+\sqrt{2-u^2}}2 \quad \text{and} \quad \cos t=\frac{u-\sqrt{2-u^2}}2$$

Answer 4

Substitute $x=\frac{1-y}{1+y} $ instead

\begin{align} \int\frac{x^3+3x+2}{(x^2+1)^2(x+1)}\overset{}{dx} =&\ \frac12\int \frac{y^3-3y^2-3y-3}{(1+ y^2)^2}dy\\ =& \ \frac12\int \frac{(y^2-3)y}{(1+ y^2)^2}-\frac3{1+ y^2}\ dy\\ =& \ \frac1{1+ y^2}+\frac14\ln(1+ y^2)-\frac32\tan^{-1}y \end{align}