in the first example you can actually just apply the DI method on the integral of (lnx)^2 by differentiating (lnx)^2 and integrating 1: D I +(lnx)^2 1 -(2lnx)/x x = x(lnx)^2 - Integral of 2lnx = x(lnx)^2 - 2xlnx + 2x + C
Can u solve integration -a to a [ f(x) ] / [ 1 + b^( g(x) )] dx =1/2 [ integration -a to a f(x) dx ] prove that if f(x) is even function and g(x) is odd function
in the first example you can actually just apply the DI method on the integral of (lnx)^2 by differentiating (lnx)^2 and integrating 1:
D I
+(lnx)^2 1
-(2lnx)/x x
= x(lnx)^2 - Integral of 2lnx
= x(lnx)^2 - 2xlnx + 2x + C
I feel like I've gained superpowers from watching these videos.
hiii , can you explain more about integral of (ln(x))^2 , how did you get dx=e^u du?
is it because u use implicit differenitation?
Since u=ln(x)
so x=e^u by the inverse property
the diff both sides
we have dx=e^u du
@@bprpcalculusbasics thankyouuuuuu! :))
Can u solve
integration -a to a [ f(x) ] / [ 1 + b^( g(x) )] dx =1/2 [ integration -a to a f(x) dx ] prove that if f(x) is even function and g(x) is odd function
I solved that already.
Can you please explain properly why do you choose certain functions to integrate and to differentiate.
lets goooooo always great
For number 4 i think it's easier to let x^2 = tan(u)
great