CUET UG 2025 — Mathematics Calculus

The integral is $\displaystyle \int \frac{\sqrt{16+(\log x)^2}}{x} dx$∫x16+(logx)2​​dx

The presence of $\frac{1}{x}$x1​ and $\log x$logx suggests a substitution approach.

Let $t = \log x$t=logx

Then $\frac{dt}{dx} = \frac{1}{x}$dxdt​=x1​

So $dt = \frac{1}{x} dx$dt=x1​dx

---

The integral becomes:

$$\int \sqrt{16+t^2} \, dt$$∫16+t2​dt

---

Using the standard formula:

$$\int \sqrt{a^2+t^2} \, dt = \frac{t}{2}\sqrt{a^2+t^2} + \frac{a^2}{2}\log\left|t + \sqrt{a^2+t^2}\right| + C$$∫a2+t2​dt=2t​a2+t2​+2a2​log​t+a2+t2​​+C

With $a^2 = 16$a2=16, so $a = 4$a=4:

$$\int \sqrt{16+t^2} \, dt = \frac{t}{2}\sqrt{16+t^2} + \frac{16}{2}\log\left|t + \sqrt{16+t^2}\right| + C$$∫16+t2​dt=2t​16+t2​+216​log​t+16+t2​​+C

$$= \frac{t}{2}\sqrt{16+t^2} + 8\log\left|t + \sqrt{16+t^2}\right| + C$$=2t​16+t2​+8log​t+16+t2​​+C

---

Substituting back $t = \log x$t=logx:

$$= \frac{\log x}{2}\sqrt{16+(\log x)^2} + 8\log\left|\log x + \sqrt{16+(\log x)^2}\right| + C$$=2logx​16+(logx)2​+8log​logx+16+(logx)2​​+C

This can be written as:

$$= 8\log\left|\log x + \sqrt{16+(\log x)^2}\right| + \frac{\log x}{2}\sqrt{16+(\log x)^2} + C$$=8log​logx+16+(logx)2​​+2logx​16+(logx)2​+C