Integration and different approaches


!!int(2(2y^2+1))/(y^2+1)^(1/2) dy!!

!!(1)" "2y(y^2+1)^(1/2)!!


!!(3)" "y^2(y^2+1)^0.5!!

!!(4)" "!!None


This question can be solved in more than one way let us look at those.

First approach:

Our main idea is to simplify this equation into easily integral form.
Denominator should be such that after its differentiation we can get it into some form of numerator. Taking this into consideration we can multiply and divide given equation with !!y!!.

!!int(2(2y^2+1))/(y^2+1)^(1/2) dy!!

!!= int(2y(2y^2+1))/(y(y^2+1)^(1/2) dy!!

!!= int(4y^3+2y)/(y^2(y^2+1))^(1/2) dy!!

!!= int(4y^3+2y)/(y^4+y^2)^(1/2) dy!!.............(i)

Now if we substitute !!y^4+y^2 = z!!......(ii) and differentiating both LHS and RHS with respect to y.

We get, !!(4y^3+2y)dy = dz!!

We can see that this equation is equivalent to numerator of equation (i).

Hence, now equation (i) can be written as:

!!= int(1/sqrtz) dz!!

After integration:

!!= 2sqrtz!!

Now substituting value of !!z!! from equation (ii), we get

!!= 2sqrt(y^4+y^2)!!

!!= 2ysqrt(y^2+1)!!

Second approach:

In the second approach we can go for converting numerator in the form of denominator.

!!int(2(2y^2+1))/(y^2+1)^(1/2) dy!!

!!= int(4y^2+2)/(y^2+1)^(1/2) dy!!

By adding and substracting 2 at the numerator, we get

!!= int(4y^2+2+2-2)/(y^2+1)^(1/2) dy!!

!!= int(4y^2+4-2)/(y^2+1)^(1/2) dy!!

!!= int(4(y^2+1)-2)/(y^2+1)^(1/2) dy!!

!!= int4(y^2+1)/(y^2+1)^(1/2) dy" "-" "int2/(y^2+1)^(1/2) dy!!

!!= int4s(y^2+1)^(1/2) dy" "-" "int2/(y^2+1)^(1/2) dy!!

Integrating this, we get

!!= 4((y/2)sqrt(y^2+1)" "+" "(1/2)log(y+sqrt(y^2+1)))" "-" "2log(y+sqrt(y^2+1)!!

!!= 2ysqrt(y^2+1)" "+" "2log(y+sqrt(y^2+1)))" "-" "2log(y+sqrt(y^2+1)!!

!!= 2ysqrt(y^2+1)!!

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