CUET Preparation Today
$\int(e^{x\log a}+e^{a\log x)} dx$ is equal to (where $a > 1$) |
$\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+C$: C is an arbitrary constant $(\log a) a^x+\frac{x^{a+1}}{a+1}+C$: C is an arbitrary constant $\frac{a^x}{a+1}+\frac{x^{a+1}}{\log a}+C$: C is an arbitrary constant $(a+1)a^x+(\log a)x^{a+1}+C$: C is an arbitrary constant |
$\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+C$: C is an arbitrary constant |
The correct answer is Option (1) → $\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+C$: C is an arbitrary constant $\int (e^{x\log a} + e^{a\log x})\,dx$ $= \int a^{x}\,dx + \int x^{a}\,dx$ $= \frac{a^{x}}{\log a} + \frac{x^{a+1}}{a+1} + C$ $\frac{a^{x}}{\log a} + \frac{x^{a+1}}{a+1} + C$ |
