# 3 integration technique problems

• Nov 9th 2010, 06:46 PM
gbux512
3 integration technique problems
the first is a division problem,

int of 3x^4 + 3x^3 - 5x^2 + x - 1 / x^2 + x - 2 dx
3x + 0 - 11
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i started by dividing it and getting x^2 + x - 2 | 3x^4 + 3x^3 - 5x^2 x - 1
-(3x^4 + 3x^3 - 6x^2)
- 11x^2 + x - 1
-(-11x^2 + 11x +22)
12x + 23

i dont exactly remember long division, but thats where im a so far, do i now sub that 3x-11 thing back into the integral? what about the remainder

the second problem is
int x^3 + 3x^2 + x - 9/ (x^2 + 1)(x^2 + 3) dx

is this decomposing fractions? would it be x^3 + 3x^2 + x - 9= Ax+b/(x^2 + 3) and cx + d/(x^2 + 1)
im completely lost on this guy

the third problem, i thing i made the most progress on

int dx/ sqrt x - 4throot x

then u ^4= x
then 4u^3 du = dx

sub it all back in
int 4u^3 du/ (u^2) ^1/4 - (u^4)^1/2

i dont know where to go from here.

I really could use some help with these. Im not looking for an easy answer, just set me on the right path and ill be grateful :)
• Nov 9th 2010, 07:12 PM
pickslides
Quote:

Originally Posted by gbux512
the third problem, i thing i made the most progress on

int dx/ sqrt x - 4throot x

$\displaystyle \displaystyle \int \frac{dx}{\sqrt{x}- \sqrt[4]{x}}$
make $\displaystyle u = \sqrt[4]{x} \implies du = \frac{1}{4x^{\frac{3}{4}}}$
Giving $\displaystyle \displaystyle 4\int \frac{u^3}{u^2-u}~du = 4\int \frac{u^2}{u-1}~du$