Question:
What is \(\int \frac{dx}{x(x^7 + 1)}\) equal to?
\(\int \frac{dx}{x(x^7 + 1)}\) किसके बराबर है?
Options:
- \(\frac{1}{7}\ln\left|\frac{x^7 – 1}{x^7}\right| + C\)
- \(\frac{1}{7}\ln\left|\frac{x^7 + 1}{x^7}\right| + C\)
- \(\ln\left|\frac{x^7 – 1}{7x^7}\right| + C\)
- \(\frac{1}{7}\ln\left|\frac{x^7}{x^7 + 1}\right| + C\)
- \(\frac{1}{7}\ln\left|\frac{x^7 – 1}{x^7}\right| + C\)
- \(\frac{1}{7}\ln\left|\frac{x^7 + 1}{x^7}\right| + C\)
- \(\ln\left|\frac{x^7 – 1}{7x^7}\right| + C\)
- \(\frac{1}{7}\ln\left|\frac{x^7}{x^7 + 1}\right| + C\)
Correct Answer: d)
Explanation:
Multiply numerator and denominator by \(x^6\):
\[\int \frac{x^6}{x^7(x^7 + 1)}\,dx\]
Let \(t = x^7\), then \(dt = 7x^6\,dx \implies x^6\,dx = \frac{dt}{7}\).
The integral becomes:
\[\frac{1}{7}\int \frac{dt}{t(t+1)} = \frac{1}{7}\int \left(\frac{1}{t} – \frac{1}{t+1}\right) dt = \frac{1}{7}[\ln|t| – \ln|t+1|] + C = \frac{1}{7}\ln\left|\frac{x^7}{x^7 + 1}\right| + C\]
Hence, the correct option is (d).
अंश और हर को \(x^6\) से गुणा करने पर:
\[\int \frac{x^6}{x^7(x^7 + 1)}\,dx\]
माना \(t = x^7\), तो \(dt = 7x^6\,dx\):
\[\frac{1}{7}\int \left(\frac{1}{t} – \frac{1}{t+1}\right) dt = \frac{1}{7}\ln\left|\frac{x^7}{x^7 + 1}\right| + C\]
अतः सही विकल्प (d) है।