Integrate (4x^2-4x +7)/(4x^2-4x +3)?

Question

Thankyou

Answer 1

∫ [(4x² - 4x + 7) /(4x²- 4x + 3)] dx =

since the denominator and the numerator are of the same degree (whereas the degree of the denominator should be lower), reduce the degree of the denominator as follows:

rewrite the numerator as:

∫ {[4x² - 4x + (4 + 3)] /(4x²- 4x + 3)} dx =

∫ {[(4x² - 4x + 3) + 4] /(4x²- 4x + 3)} dx =

split it as:

∫ {[(4x² - 4x + 3)/(4x²- 4x + 3)] + [4 /(4x²- 4x + 3)] } dx =

∫ [(4x² - 4x + 3)/(4x²- 4x + 3)] dx + ∫ [4 /(4x²- 4x + 3)] dx =

simplifying:

∫ dx + ∫ [4 /(4x²- 4x + 3)] dx =

x + ∫ [4 /(4x²- 4x + 3)] dx (#) =

as for the remaining integral, note that its denominator is not factorable,

thus you have to complete the square as:

(4x²- 4x + 3) =

4x²- 4x + (1 + 2) =

(4x²- 4x + 1) + 2 =

(2x - 1)² + 2 =

then, in order to rearrange the denominator into {[f(x)]² + 1} form,

factor out 2:

2 [(1/2)(2x - 1)² + 1] =

then include (1/2) into the square, yielding:

2 {[(1/√2)(2x - 1)]² + 1} =

expand the square base into:

2 { [x√2 - (1/√2)]² + 1} =

now, going back to (#) step above, and plugging in the

rearranged integrand, you get

x + ∫ [4 /(4x²- 4x + 3)] dx =

x + ∫ 4 dx / {2 { [x√2 - (1/√2)]² + 1}} =

x + ∫ 2 dx / { [x√2 - (1/√2)]² + 1} =

finally, in order to turn the numerator into the derivative of [x√2 - (1/√2)]

(i.e.√2), split it into √2 √2 and factor √2 out, yielding:

x + ∫ √2 √2 dx / { [x√2 - (1/√2)]² + 1} =

x + √2 ∫ √2 dx / { [x√2 - (1/√2)]² + 1} =

x + √2 ∫ {d[x√2 - (1/√2)]} / {[x√2 - (1/√2)]² + 1} =

x + √2 arctan[x√2 - (1/√2)] + C

in conclusion:

∫ [(4x² - 4x + 7) /(4x²- 4x + 3)] dx = x + √2 arctan[x√2 - (1/√2)] + C

I hope this helps...

Bye!

Answer 2

x+ilog(-2ix+squareroot2+i/squarerootof2-...

Answer 3

Use Polynomial Division...

................___1_______

4x^2-4x+3 | 4x^2 -4x +7

.................-(4x^2-4x+3)

..................0        0     4

the answer is, 1 + (4 / 4x^2 -4x +3)

Answer 4

=   (4/3 x^3 - 4/2 x^2 + 7x) / (4/3 x^3 - 4/2 x^2 +3x)

=  (4/3 x^3 - 2x^2 + 7x) / (4/3 x^3 - 2x^2 +3x)

Answer 5

use polynomial division as a first step:-

(4x^2 - 4x + 7)/(4x^2 - 4x + 3)  =  1  +  (4 / (4x^2 - 4x + 3))

and then you can integrate each term on the right, the '1' is very easy, and the other part you should be able to deduce from a table of standard integrals after having divided top and bottom by '4'.

i make this part to be:-

Integral (4 / 4x^2 - 4x + 3)  =  (sqrt(2)x - sqrt(2)/2)*atan(sqrt(2)x - sqrt(2)/2) - 1/2ln[1 + (sqrt(2)x - sqrt(2)/2)^2]

Answer 6

I think you can use a basic mathematical formula with this one

solve each equation within the brackets and then divide

4x cancels out -4x

^2

+7

divided by

4x cancels out -4x

leaving

^2

+3

now you have

(^2+7)/(^2+3)=

that's has far as I can guess.