Q80 / 170 · Manipal (MET) 2017 · Mathematics

$\int\left(e^{a \log x}+e^{x \log a}\right) d x$ is equal to

  • A
    $\frac{x^{a+1}}{a+1}+c$
  • Correct answer
    B
    $\frac{x^{a+1}}{a+1}+\frac{a^x}{\log a}+c$
  • C
    $x^{a+1}+a^x+c$
  • D
    $\frac{x^{a+1}}{a-1}+\frac{\log a}{a^x}+c$

Solution

$\begin{aligned} & \int\left(e^{a \log x}+e^{x \log a}\right) d x=\int\left(e^{\log x^a}+e^{\log a^x}\right) d x \\ = & \int\left(x^a+a^x\right) d x=\frac{x^{a+1}}{a+1}+\frac{a^x}{\log _e a}+c\end{aligned}$

Hence, the correct answer is option: 'B'.

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