\(\int\displaystyle\frac{\pi}{x^{n+1}-x}\,dx=\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\pi}{n}\log_e\left|\frac{x^n-1}{x^n}\right|+C\)
Factorising the denominator, \(x^{n+1}-x = x\left(x^n-1\right)\). Multiplying the numerator and the denominator by \(x^{n-1}\):
\( \displaystyle I = \int\frac{\pi}{x\left(x^n-1\right)}\,dx = \pi\int\frac{x^{n-1}}{x^n\left(x^n-1\right)}\,dx \)
Now the substitution \(t = x^n\) works, since \(dt = nx^{n-1}\,dx\):
\( \displaystyle I = \frac{\pi}{n}\int\frac{dt}{t(t-1)} \)
By partial fractions:
\( \displaystyle \frac{1}{t(t-1)} = \frac{1}{t-1}-\frac{1}{t} \)
So
\( \displaystyle I = \frac{\pi}{n}\left(\log_e|t-1|-\log_e|t|\right)+C \)
\( \displaystyle = \frac{\pi}{n}\log_e\left|\frac{t-1}{t}\right|+C \)
\( \displaystyle = \frac{\pi}{n}\log_e\left|\frac{x^n-1}{x^n}\right|+C \)