In this video, we will learn to find integral
of x upon square root of x plus 1.
Let I be equal to integral of x upon square
root of x plus 1 w.r.t. x.
We know adding a constant and subtracting
the same constant to an expression does not
change the value of the expression. So, I
can be re-written as integral of x plus 1
minus 1 upon square root of x plus 1 w.r.t.
x.
The integrand can be expressed as the difference
of two fractions so, I is equal to integral
of x plus 1 upon square root of x plus 1 minus
1 upon square root of x plus 1 w.r.t. x.
Canceling out the common factor square root
of x plus 1 in the numerator and the denominator
of first fraction of the above integrand gives
us I is equal to integral of square root of
x plus 1 minus 1 upon square root of x plus
1 w.r.t. x.
In exponent form it can be written as I is
equal to integral of x plus 1 raised to the
power of half minus x plus 1 raised to the
power of minus half w.r.t. x.
Let’s further simplify the integral by application
of substitution method. Assume x plus 1 is
equal to t.
Let’s find the value of differential of
t, taking differential of both sides of the
above equation gives us, derivative of x plus
1 w.r.t. x times differential of x is equal
to differential of t.
We know derivative of x is 1 and derivative
of constant 1 is 0. So, differential of x
is equal to differential of t.
In the integral I, let’s substitute x plus
1 by our assumed value t and differential
of x by differential of t. So, we get, I is
equal to integral of t raised to the power
of half minus t raised to the power of minus
half w.r.t. t.
We know integral of t raised to the power
of n is equal to t raised to the power of
n plus 1 upon n plus 1. So, I is equal to
t raised to the power of half plus 1 upon
half plus 1 minus t raised to the power of
minus half plus 1 upon minus half plus 1 plus
capital c. Capital c is the arbitrary constant
of indefinite integral. Never forget to add
this constant.
As, half plus 1 is equal to 3 upon 2 and minus
half plus 1 is equal to half. So, I is equal
to t raised to the power of 3 upon 2 upon
3 upon 2 minus t raised to the power of half
upon half plus capital c.
Further simplification gives us I is equal
to 2 times t raised to the power of 3 upon
2 upon 3 minus 2 times t raised to the power
of half plus capital c.
Let’s substitute t in terms of x. As we
had assumed t is equal to x plus 1. So, I
is equal to 2 times x plus 1 raised to the
power of 3 upon 2 upon 3 minus 2 times x plus
1 raised to the power of half plus capital
c.
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