Integration by Substitution (includes Examples) Video Lecture | Mathematics (Maths) Class 12 - JEE

in this tutorial I'm going to discuss
integration by substitution or called u
substitution I'm going to find the
integral of this function and the final
answer looks like this don't run
hang in there on my walk you through
step by step
if you look at this problem you'll find
that there's a relationship between x
squared plus 1 and 4x and in fact if I
take the derivative of x squared plus 1
it's kind of like 4 X so the derivative
of x squared plus 1 is 2x and if you
watch a couple of these examples you'll
kind of get the hang of this so now let
me do a use substitution so U is equal
to x squared plus 1 the derivative of U
is d u so if I take the derivative of x
squared that's 2x plus the derivative of
1 is 0 the derivative of any constant is
a 0 so now D U is equal to 2x times DX
I'm going to solve for DX I'm going to
put it right there I multiply both sides
of the equation by 1 over 2x and notice
on the right hand side these two x's
cancel out now I have D u over 2x is
equal to DX and I'm going to put it
right there in a moment let me copy this
4x divided by the square root of x
squared plus 1 over right where this x
squared plus 1 is right there I'm going
to put you like that and where there's
DX I'm going to put D u over 2x now if
you notice this 2x and is for X the X is
canceled out and I'm left with 2 divided
by the square root of U D u the
looks like something I can actually
integrate now I'm going to move this up
and give myself a little bit more space
I'm going to factor out the two and this
square root of U is equal to U to the
one-half power so I have two times the
integral of 1 divided by U to the
one-half power D u I can also write this
as 2 times the integral of 1 times U to
the negative 1/2 power D you now I'm
going to take the antiderivative of this
so I have 2 times 1 times u raised to
the negative 1/2 power plus one like
that I'm adding 1 to the exponent
times 1 is just 2 negative 1/2 plus 1 is
1/2 and when I take an antiderivative
I've take this power this new power and
divide by that again so it's divided by
1/2 2 divided by 1/2 2 divided by 1/2 is
4 now I have 4 times U to the one-half
power and if I put the you back end
which is x squared plus 1 plus some
constant don't forget your C plus a C
that's my answer right there
get rid of everything else and you'll
see there we go now I'm going to check
my answers should always check your
answer by taking the derivative of this
bad boy down here the first thing you
can use is lob off that C because the
derivative of any constant is nothing
zero now I use the chain rule I take the
one-half power and multiply it in front
of everything so I move it in the front
times everything and I subtract one now
I take the derivative of the inside
which is 2 X 1/2 times 4 is 2 the new
exponent or the power of x squared plus
1 is negative 1/2 and I just pull this
2x down I continue 2 times 2x is 4x this
is looking up so now I have 4x times x
squared plus 1 to the negative 1/2 power
now I have 4x divided by the square root
of x squared plus 1 anything raised to
the one-half power is the square root of
and since it's a negative I put it on
the denominator so look at that so I'm
talking about right here that's the
right answer and here you go here's the
complete problem worked out the
integration of it and remember to share
the knowledge should live Facebook
Google+ and Twitter links to other
videos examples below and subscribe and
please like me please I don't have any
friends either
please please please