Question

$\int \frac{1}{x\left(x^5+1\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{1}{x\left(x^5+1\right)}dx \\ =\int \frac{x^{4}dx}{x^5\left(x^5+1\right)} \\ \mathrm{Putting}{x}^5=t \\ \Rightarrow 5x^{4}dx=dt \\ \Rightarrow x^{4}dx=\frac{dt}{5} \\ \therefore I=\frac{1}{5}\int \frac{dt}{t\left(t+1\right)}.....\left(1\right) \\ \mathrm{Let}\frac{1}{t\left(t+1\right)}=\frac{A}{t}+\frac{B}{t+1}.....\left(2\right) \\ \Rightarrow \frac{1}{t\left(t+1\right)}=\frac{A\left(t+1\right)+Bt}{t\left(t+1\right)} \\ \Rightarrow 1=A\left(t+1\right)+Bt.....\left(3\right) \\ \mathrm{Putting}t=0\mathrm{in}\left(3\right) \\ \Rightarrow 1=A\left(0+1\right)+B\times 0 \\ \Rightarrow A=1 \\ \mathrm{Putting}t+1=0\mathrm{in}\left(3\right) \\ \Rightarrow t=-1 \\ \therefore 1=A\times 0+B\left(-1\right) \\ \Rightarrow B=-1 \\ \therefore I=\frac{1}{5}\left[\int \frac{dt}{t}-\int \frac{dt}{t+1}\right]\left[\mathrm{From}\left(1\right)\&\left(2\right)\right] \\ =\frac{1}{5}\left[\log \left|t\right|-\log \left|t+1\right|\right]+C \\ =\frac{1}{5}\log \left|\frac{t}{t+1}\right|+C \\ =\frac{1}{5}\log \left|\frac{x^5}{x^5+1}\right|+C \end{gathered}$$