Hello, soni!
Your difficulty is understandable.
$\displaystyle \displaystyle \text{Evaluate: }\;\int^1_0\sqrt{1+16x^2}\,dx$
Let $\displaystyle 4x \,=\,\tan\theta \quad\Rightarrow\quad dx \:=\:\frac{1}{4}\sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{1+16x^2} \:=\:\sec\theta $
$\displaystyle \displaystyle\text{Substitute: }\:\int \sec\theta\cdot\tfrac{1}{4}\sec^2\!\theta\,d\theta \;=\;\tfrac{1}{4}\int \sec^3\!\theta\,d\theta$
And we have the dreaded secant-cubed integral . . .
The formula can be derived by integrating-by-parts twice.
Or we can simply memorize the formula:
. . $\displaystyle \displaystyle \int\sec^3\!\alpha\,dx \:=\:\tfrac{1}{2}\bigg[\sec\alpha \tan\alpha + \ln\left|\sec\alpha + \tan\alpha\right|\bigg] + C$
So we have: .$\displaystyle \frac{1}{4}\cdot\frac{1}{2}\bigg[\sec\theta\tan\theta + \ln\left|\sec\theta + \tan\theta\right|\bigg] + C $
Back-substitute: .$\displaystyle \tan\theta = 4x,\;\sec\theta = \sqrt{1+16x^2} $
And we have: .$\displaystyle \frac{1}{8}\bigg[4x\sqrt{1+16x^2} + \ln\left|\sqrt{1+16x^2} + 4x\right|\bigg]^1_0$
Got it?
What I would have done after reaching $\displaystyle \frac{1}{4}\int sec^3(\theta)d\theta$
Rewrite it as $\displaystyle \frac{1}4}\int\frac{d\theta}{cos^3(\theta)}$
Since that is now an odd power of $\displaystyle cos(\theta)$, multiply both numerator and denominator by $\displaystyle cos(\theta)$ to get
$\displaystyle \frac{1}{4}\int\frac{cos(\theta)d\theta}{cos^4(\th eta)}= \frac{1}{4}\int\frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}$
and make the substitution $\displaystyle u= sin(\theta)$:
$\displaystyle \frac{1}{4}\int\frac{du}{(1- u^2)^2}= \frac{1}{4}\int\frac{du}{(1- u)^2(1+ u)^2}$
a rational integral. It can be done by partial fractions:
$\displaystyle \frac{1}{(1- u)^2(1+ u)^2}= \frac{A}{1- u}+ \frac{B}{(1- u)^2}+ \frac{C}{1+ u}+ \frac{D}{(1+ u)^2}$
Eventually giving the same result Soroban got.