CUET UG 2025 — Mathematics Calculus

Given: $\displaystyle \int (e^{x\log a} + e^{a\log x}) dx$∫(exloga+ealogx)dx

For the first term $e^{x\log a}$exloga:

Using the property $n\log b = \log b^n$nlogb=logbn:

$$x\log a = \log a^x$$xloga=logax

Therefore:

$$e^{x\log a} = e^{\log a^x} = a^x$$exloga=elogax=ax

---

For the second term $e^{a\log x}$ealogx:

$$a\log x = \log x^a$$alogx=logxa

Therefore:

$$e^{a\log x} = e^{\log x^a} = x^a$$ealogx=elogxa=xa

---

The integral simplifies to:

$$\int (a^x + x^a) dx$$∫(ax+xa)dx

---

For $\displaystyle \int a^x dx$∫axdx:

$$\int a^x dx = \frac{a^x}{\log a} + C_1$$∫axdx=logaax​+C1​

---

For $\displaystyle \int x^a dx$∫xadx:

$$\int x^a dx = \frac{x^{a+1}}{a+1} + C_2$$∫xadx=a+1xa+1​+C2​

---

Combining the results:

$$\int (a^x + x^a) dx = \frac{a^x}{\log a} + \frac{x^{a+1}}{a+1} + C$$∫(ax+xa)dx=logaax​+a+1xa+1​+C

where $C$C is an arbitrary constant.