# Thread: Integration

1. ## Integration

I can't seem to get the answer to

2xy^3/(x^2y^2+1)

2. Please mention if it is

$\displaystyle \int \dfrac{2xy^3}{x^2y^2+1} \;dx$

or

$\displaystyle \int \dfrac{2xy^3}{x^2y^2+1} \;dy$

and show your attempt.

3. Sorry it is dy

I tried a substitution and it got very messy, ended up with y= sqrt( u-1/x). Hence I think I have taken the completely wrong route

4. $\displaystyle \int \dfrac{2xy^3}{x^2y^2+1} \;dy$

$= \displaystyle 2x \; \int \dfrac{y^3}{x^2y^2+1} \;dy$

Use Partial fractions:

$\dfrac{y^3}{x^2y^2+1} = \dfrac{y}{x^2} - \dfrac{y}{x^2 (x^2 y^2+1)} dy$

5. Or you may make the substitution $x^{2}y^{2} = t$,

so that $dt = 2x^{2}ydy$

Therefore $dy = dt/(2x^{2}y)$

and $y^{2} = t/x^{2}$
Therefore the integrand now becomes, $\int(1/x^{3})(t/(t+1))\;dt$

This can be evaluated using standard forms.

6. Thank you

I can integrate y/x^2 dy to get y^2/2x^2

Do I tackle the second integral by substitution?

I got du=2x^4y dy and my result for y again seems wrong as I got square root of -1/x^2 which you can't have

7. $\displaystyle \int \dfrac{y^3}{x^2y^2+1} =\displaystyle \int 2x \dfrac{y}{x^2}dy - \displaystyle \int 2x \dfrac{y}{x^2 (x^2 y^2+1)} dy$

$=\displaystyle \int 2 \dfrac{y}{x} \;dy -\displaystyle \int 2 \dfrac{y}{x(x^2 y^2+1)} dy$

for the second one, use substution.

$u=x^2y^2+1 \; \therefore du = x^2(2y)\;dy$ and finish..