Question:
What is \(\int \frac{x^{e-1} + e^{x-1}}{x^e + e^x}\,dx\) equal to?
\(\int \frac{x^{e-1} + e^{x-1}}{x^e + e^x}\,dx\) किसके बराबर है?
Options:
- \(\frac{x^2}{2} + x\)
- \(\ln(x + e) + c\)
- \(\ln(x^e + e^x) + c\)
- \(\frac{1}{e}\ln(x^e + e^x) + c\)
- \(\frac{x^2}{2} + x\)
- \(\ln(x + e) + c\)
- \(\ln(x^e + e^x) + c\)
- \(\frac{1}{e}\ln(x^e + e^x) + c\)
Correct Answer: d)
Explanation:
Let \(t = x^e + e^x\).
Differentiating with respect to \(x\):
\[\frac{dt}{dx} = e x^{e-1} + e^x = e(x^{e-1} + e^{x-1}) \implies (x^{e-1} + e^{x-1})\,dx = \frac{dt}{e}\]
Substituting into the integral:
\[\int \frac{1}{t} \cdot \frac{dt}{e} = \frac{1}{e}\ln|t| + c = \frac{1}{e}\ln(x^e + e^x) + c\]
Hence, the correct option is (d).
माना \(t = x^e + e^x\)।
अवकलन करने पर:
\[dt = (e x^{e-1} + e^x)\,dx = e(x^{e-1} + e^{x-1})\,dx \implies (x^{e-1} + e^{x-1})\,dx = \frac{dt}{e}\]
समाकलन करने पर:
\[\frac{1}{e}\int \frac{dt}{t} = \frac{1}{e}\ln(x^e + e^x) + c\]
अतः सही विकल्प (d) है।