Question

$\int \left[e^{a\log x}+e^{x\log a}\right]dx=\dots ..+c,a,x>1$

• A. $\left[\frac{x^{a+1}}{(a+1)}\right]+\left(\frac{a^x}{\log a}\right)$
• B. $\left[\frac{e^{a\log x}}{a\log x}\right]+\left[\frac{e^{x\log a}}{x\log a}\right]$
• C. $\left[\frac{x^{a-1}}{a–1}\right]+a^x.\log a+c$
• D. $\left[\frac{e^{a\log x}}{{\displaystyle \frac{a}{x}}}\right]+\left[\frac{e^{x\log a}}{x}\right]$

Answer 1

The correct option is A

$\left[\frac{x^{a+1}}{(a+1)}\right]+\left(\frac{a^x}{\log a}\right)$

Explanation for the correct option:

Evaluating the integral:

Given

$$\begin{gathered} \int \left(e^{a\log x}+e^{x\log a}\right)dx \\ =\int (x^a+e^{x\log a})dx\left[\because e^{n\log m}=m^n\right] \\ =\left[\frac{x^{a+1}}{(a+1)}\right]+\left(\frac{e^{x\log a}}{\log a}\right)+c \\ =\left[\frac{x^{a+1}}{a+1}\right]+\left[\frac{a^x}{\log a}\right]+c \end{gathered}$$

Hence, the correct answer is option (A).