someone gud wit math plz help...int ( y+2/(2y^2+3y+1) dy )?

Question

If u can show me a few steps how u got to the answer i would appreciate it ..

Answer 1

Rule of thumb: If the degree of the numerator is less than the degree of the denominator, you have to use partial fractions. First, factor the denominator. Through trial and error, you'll find that

2y^2 + 3y + 1 = (2y + 1) (y + 1)

For now, we're going to ignore the integral, because what we're concerned about right now is the form of the function to integrate. Therefore

(y+2)/(2y^2+3y+1) = (y+2) /[ (2y+1)(y+1) ]

Through the knowledge of partial fractions,

(y+2) / [ (2y + 1) (y+1) ] = A/(2y+1) + B/(y+1), for some constants A and B. Our goal is to determine what A and B are.

Multiplying both sides by the denominator (2y+1)(y+1), we get

y + 2 = A(y+1) + B(2y+1)

We know that this is true *for all y*. So we can solve for A and B using any value for y. Let's choose y = -1 and then choose y=-1/2.

y = -1:

-1+2 = A(-1+1) + B(2(-1)+1)
1 = A(0) + B(-1)
1 = -B
Therefore, B = -1

y = -1/2:
-1/2 + 2 = A(-1/2 + 1) + B(2(-1/2) + 1)
3/2 = A(1/2) + B(0)
3/2 = A/2
Therefore, A = 3

So now we know how to split up that one big fraction into two. That is,

(y+2) / [ (2y + 1) (y+1) ] = 3/(2y+1) + -1/(y+1). So now, we integrate the new fraction.

Integral ( 3/(2y+1) + -1/(y+1) )dy = (split into two integrals and pull out constants in the process).

3*Integral(1/(2y+1))dy - Integral(1/(y+1))dy =

3*1/2*ln|2y+1| - ln|y+1| + C

3/2 * ln|2y+1| - ln|y+1| + C

Reminder: simple substitution will reveal that Integral(1/(2y+1)) = 1/2*ln|2y+1|, hence the above steps.

The above can further be simplified using rules for adding and subtracting logs, as well as the power rule for logs, but I'll leave it for you to figure out, as the more important steps were shown above.

Answer 2

note that 2y^2+3y+1 can be factored into (y+1)(2y+1)

so ( y+2/(2y^2+3y+1) = (y+2) / (y+1)(2y+1) which can be written in the form of:

A/(y+1) + B/(2y+1)

Equating both sides, we can determine A and B, being constants

Then apply the formula : int(c/x)dx = clog x + c'

Answer 3

y + 2/(2y+1)(y+1) = y + 2[2/(2y+1) - 1/(y+1)]

So, integral is y^2/2 + 2ln[(2y+1)/(y+1)]

Answer 4

Answer is
3Y^ 6Y x DY

Answer 5

Hun, with that spelling, you need someone who is good with English!