Concept:

Integration by parts: Integration by parts is a method to find integrals of products. The formula for integrating by parts is given by:

⇒ \(\rm ∫ u vdx=u ∫ vdx- ∫ \left({du\over dx}\times \int vdx\right)dx \) + C

where 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 + e^x = t²            .... (1)

Differentiating with respect to x, we get

⇒ e^x dx = 2tdt

From equation (1), we get

e^x = t² - 1

So, 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) dt

Using 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) \(\rm \sqrt{1+e^x}\) - \( \rm 2 \log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\)

= \(\rm (2x-4)\sqrt{1+e^x}- \rm 2 \log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\)