Question

$\int \frac{1}{x\left(x^4+1\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{x\left(x^4+1\right)} \\ =\int \frac{x^{3}dx}{x^4\left(x^4+1\right)} \\ \mathrm{Putting}{x}^4=t \\ \Rightarrow 4x^{3}dx=dt \\ \Rightarrow x^{3}dx=\frac{dt}{4} \\ \therefore I=\frac{1}{4}\int \frac{dt}{t\left(t+1\right)} \\ \mathrm{Let}\frac{1}{t\left(t+1\right)}=\frac{A}{t}+\frac{B}{t+1} \\ \Rightarrow \frac{1}{t\left(t+1\right)}=\frac{A\left(t+1\right)+Bt}{t\left(t+1\right)} \\ \Rightarrow 1=A\left(t+1\right)+Bt \\ \mathrm{Putting}t+1=0 \\ \Rightarrow t=-1 \\ \therefore 1=\mathrm{A}\times 0+\mathrm{B}\left(-1\right) \\ \Rightarrow \mathrm{B}=-1 \\ \mathrm{Putting}t=0 \\ \therefore 1=\mathrm{A}\left(1\right)+B\times 0 \\ \Rightarrow \mathrm{A}=1 \\ \therefore I=\frac{1}{4}\int \frac{dt}{t}-\frac{1}{4}\int \frac{dt}{t+1} \\ =\frac{1}{4}\log \left|t\right|-\frac{1}{4}\log \left|t+1\right|+\mathrm{C} \\ =\frac{1}{4}\log \left|\frac{t}{t+1}\right|+\mathrm{C} \\ =\frac{1}{4}\log \left|\frac{x^4}{x^4+1}\right|+\mathrm{C} \end{gathered}$$