INTEGRATING by parts is given by:⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times \int vdx\right)dx \) + Cwhere u is the function u(x) and v is the function v(x) ILATE rule is USUALLY, the PREFERENCE order of this rule is based on some functions such as INVERSE, Logarithm, Algebraic, Trigonometric and Exponent.Formula:\(\rm \int \frac{1}{x^2 -a^2}dx = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + c\) Calculation:Let I = \(\rm \displaystyle∫\dfrac{xe^x}{\sqrt{1+e^x}}dx\)Take 1 + ex = t2 .... (1)Differentiating with respect to x, we get⇒ ex dx = 2tdtFrom equation (1), we getex = t2 - 1So, x = log (t2 - 1)Now,I = \(\rm \displaystyle∫\dfrac{\log (t^2 - 1)}{\sqrt{t^2}}2tdt\)= \(\rm 2 × \displaystyle∫\dfrac{\log (t^2 - 1)}{t} × tdt\)= 2 ∫ log (t2 - 1) dtUsing integration by parts rule, we get= 2 [log (t2 - 1) × t - 2 \(\rm \int \frac {t^2}{t^2-1}dt\) ]= 2t log (t2 - 1) - 4 \(\rm \int \left[1+ \frac{1}{t^2-1} \right ]dt\)= 2t log (t2 - 1) - 4t - 4 × \(\rm \frac{1}{2} \log \left(\frac{t-1}{t+1} \right) +c\)= 2t log (t2 - 1) - 4t - \(\rm2\log \left(\frac{t-1}{t+1} \right) +c\)= 2t(log (t2 - 1) - 2) - \(\rm2\log \left(\frac{t-1}{t+1} \right) +c\)Resubstitute the value of t, we get= 2 (x - 2) \(\rm \sqrt{1+e^x}\) - \( \rm 2 \log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\)= \(\rm (2x-4)\sqrt{1+e^x}- \rm 2 \log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\)