Question 64

Question 62

For 1 just take common x+1 after solving

Then u will get a^2-b^2 type identity try it u will get result..

#2 looking like problem based on 2 classic integral basically by part type problm..

e^{f(x)} \left(xf'(x)+1\right)dx=d(xe^{f(x)})

2 Likes

For 1st log does not exist for negative numbers.

For first lebniz theoram but by uv rule of diffrenciation

#2 use

e^{(f(x)-1)g(x)} concept = 1^{\infty} form