# Thread: help with these integrals (i beleive they should be mostly partial fractions)

1. ## help with these integrals (i beleive they should be mostly partial fractions)

ive gotten the easy ones, but these ones are throwing me for a loop...

we got these...
integrate

(x-3)dx/(x^3+3x^2+2x)

x^2 dx(x^2+2x+2)^2

(sin^2 x) dx / (1 + sin^2 x)

2. Hint for the last one, if you're really up to integrating such a thing:

$\frac{\sin^2x}{1+\sin^2x}=1-\frac{1}{1+\sin^2x}=1-\frac{1}{2-\cos^2x}$

and now use partial fractions.

3. how were you able to rewrite that? (like what property, i dont really understand either change

4. Well for the first step, notice that

$\frac{A}{A+B}=1-\frac{B}{A+B}$.

For the second one I just used $\cos^2 x + \sin^2 x = 1$.

I can tell you that the integral is quite difficult. The answer involves both arctangent and hyperbolic arctangent!

5. the first one

$\frac{x-3}{x^3+3x^2+2x}$ = $\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x+1}$

$\frac{x-3}{x^3+3x^2+2x}$ = $\frac{(Ax^2+3Ax+2A)+(Bx^2+Bx)+(Cx^2+Cx)}{x^3+3x^2+ 2x}$ $=\frac{(A+B+C)x^2 + (3A+B+C)x + 2A}{x^3+3x^2+2x}$

solve
$(A+B+C)=0$
$(3A+B+C)=1$
$2A=-3$

6. For the second integral the intgrls2 attachment

For the first you can solve for the coefficients

by letting x = 0,-2, and -1 after clearing fractions

See intgrls3 attachment

7. can u just explain right under where it says "equating coefficents of like powers".. i understand everything else i think, i just dont get where those numbers come from

8. Two polynimials are equal if and only the coefficients of like powers are equal

The coefficients of the cubed terms are the same
the coefficients of the squared terms are the same
etc

In your example there is no cubic term on the left So the coefficient of the cubic term on the right hand side namely A must be 0

The coefficient of the squared term on the left is 1 so the coefficient
of the squared term on the right namely 2A + B =1

since A= 0 B must be one

and so on

9. thanks so much! you have no idea how much you helped me. i literally had no idea an hour ago, and now i feel pretty comfortable

10. Remember in General if you have linear factors you can easily solve for the constants by making propitious choices of x.

When you have irreducible quadratics you expand and equate coefficients
of like powers