# Thread: 3 integration technique problems

1. ## 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

2. 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 $\displaystyle \displaystyle 4\int \frac{u^3}{u^2-u}~du = 4\int \frac{u^2}{u-1}~du$