Question

$\int \frac{1}{1+x+x^2+x^3}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{1+x+x^2+x^3} \\ =\int \frac{dx}{\left(1+x\right)+x^2\left(1+x\right)} \\ =\int \frac{dx}{\left(x+1\right)\left(x^2+1\right)} \\ \mathrm{Let}\frac{1}{\left(x+1\right)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1} \\ \Rightarrow \frac{1}{\left(x+1\right)\left(x^2+1\right)}=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)} \\ \Rightarrow 1=A\left(x^2+1\right)+B{x}^2+Bx+Cx+C \\ \Rightarrow 1=\left(A+B\right)x^2+\left(B+C\right)x+\left(A+C\right) \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+B=0.....\left(1\right) \\ B+C=0.....\left(2\right) \\ A+C=1.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ A=\frac{1}{2} \\ B=-\frac{1}{2} \\ C=\frac{1}{2} \\ \therefore I=\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int \left(\frac{-x+1}{x^2+1}\right)dx \\ =\frac{1}{2}\int \frac{dx}{x+1}-\frac{1}{2}\int \frac{xdx}{x^2+1}+\frac{1}{2}\int \frac{dx}{x^2+1^2} \\ \mathrm{let}{x}^2+1=t \\ \Rightarrow 2xdx=dt \\ \Rightarrow xdx=\frac{dt}{2} \\ \therefore I=\frac{1}{2}\int \frac{dx}{x+1}-\frac{1}{4}\int \frac{dt}{t}+\frac{1}{2}\int \frac{dx}{x^2+1^2} \\ =\frac{1}{2}\log \left|x+1\right|-\frac{1}{4}\log \left|t\right|+\frac{1}{2}{\tan }^{-1}\left(x\right)+C\text{'} \\ =\frac{1}{2}\log \left|x+1\right|-\frac{1}{4}\log \left|x^2+1\right|+\frac{1}{2}{\tan }^{-1}\left(x\right)+C\text{'} \end{gathered}$$