Partial Fractions: Integrate (4x+10)/(9x^2+24x+16)

arl146
Messages
342
Reaction score
1

Homework Statement


determine the indefinite integral: ∫ (4x+10)/(9x^2+24x+16) dx


Homework Equations


partial fractions technique


The Attempt at a Solution



i know it's partial fractions and i thought i did it right but i got the wrong answer.

(4x+10)/(9x^2+24x+16) = (4x+10)/(3x+4)^2 = A/(3x+4) + B/(3x+4)^2
is that part right before I continue?
 
Physics news on Phys.org
arl146 said:

Homework Statement


determine the indefinite integral: ∫ (4x+10)/(9x^2+24x+16) dx


Homework Equations


partial fractions technique


The Attempt at a Solution



i know it's partial fractions and i thought i did it right but i got the wrong answer.

(4x+10)/(9x^2+24x+16) = (4x+10)/(3x+4)^2 = A/(3x+4) + B/(3x+4)^2
is that part right before I continue?

Yes, that is the right idea.

RGV
 
Yes, that's right.
 
arl146 said:

Homework Statement


determine the indefinite integral: ∫ (4x+10)/(9x^2+24x+16) dx


Homework Equations


partial fractions technique


The Attempt at a Solution



i know it's partial fractions and i thought i did it right but i got the wrong answer.

(4x+10)/(9x^2+24x+16) = (4x+10)/(3x+4)^2 = A/(3x+4) + B/(3x+4)^2
is that part right before I continue?
As others have told you that is correct. Now multiply both sides by (3x+4)^2 to get (4x+10)(3x+ 4)^2= A(3x+ 4)+ B. Taking x= -4/3 gives an easy solution for B. Take x to be any other number, say, x= 0, to get an equation for A.
 
ohhhh .. i see what i did. ok .. so i was doing A(3x+ 4)^2 + B(3x+4) not realizing the denominator difference. stupid mistake. thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top