Okh to solve your question I will use

i for integration and d for diffrentiation w.r.t. x okh

Apply ILATE rule

which says i= log(e^x+1) i e^xdx- i { d (log(e^x+1)) i e^xdx}dx

log(e^x+1)e^x- i {e^x/(e^x+1)*e^x}dx ( i e^x=e^x && d logx=1/x)

Now use substitution

[Let u=e^x then du=e^xdx]

The equation becomes

log(e^x+1)e^x- i {u/1+u}du

log(e^x+1)e^x- i {u+1-1/1+u}du

log(e^x+1)e^x- [ i du - i (1/1+u)du]

log(e^x+1)e^x- [u -log(1+u)] [ i ( 1/1+x)dx=log(1+x)]

Put u=e^x

log(e^x+1)e^x- [e^x -log(1+e^x)]

Hope this helps!!!

Hello aspirant,

√(tan x) dx

Let tan x = t 2

⇒ sec 2 x dx = 2t dt

⇒ dx = [2t / (1 + t 4 )]dt

⇒ Integral  ∫ 2t 2 / (1 + t 4 ) dt

⇒ ∫[(t 2 + 1) + (t 2 - 1)] / (1 + t 4 ) dt

⇒ ∫(t 2 + 1) / (1 + t 4 ) dt + ∫(t 2 - 1) / (1 + t 4 ) dt

⇒ ∫(1 + 1/t 2 ) / (t 2 + 1/t 2 ) dt + ∫(1 - 1/t 2 ) / (t 2 + 1/t 2 ) dt

⇒ ∫(1 + 1/t 2 )dt / [(t - 1/t) 2 + 2] + ∫(1 - 1/t 2 )dt / [(t + 1/t) 2 -2]

Let t - 1/t = u for the first integral ⇒ (1 + 1/t 2 )dt = du

and t + 1/t = v for the 2nd integral ⇒ (1 - 1/t 2 )dt = dv

Integral
= ∫du/(u 2 + 2) + ∫dv/(v 2 - 2)

= (1/√2) tan -1 (u/√2) + (1/2√2) log(v -√2)/(v + √2)l + c

= (1/√2) tan -1 [(t 2 - 1)/t√2] + (1/2√2) log (t 2 + 1 - t√2) / t 2 + 1 + t√2) + c

= (1/√2) tan -1 [(tanx - 1)/(√2tan x)] + (1/2√2) log [tanx + 1 - √(2tan x)] / [tan x + 1 + √(2tan x)] + c

You can check the list of colleges below

Government

Indian Institute of Technology Gandhinagar

Maharaja Sayajirao University of Baroda, Vadodara

Sardar Vallabhbhai National Institute of Technology Surat

Dharmsinh Desai University, Nadiad

LD College of Engineering, Ahmedabad

Private

Dhirubhai Ambani Institute of Information and Communication Technology, Gandhinagar

Nirma University, Ahmedabad

CEPT University, Ahmedabad

Charotar University of Science and Technology, Petlad

Pandit Deendayal Petroleum University, Gandhinagar

To check the complete list you can refer below link

https://engineering.careers360.com/colleges/ranking/2017/top-engineering-colleges-in-gujarat

Good Luck!!