integral (3+2x^2)/xdx

**khuezy** · June 6th 2008, 09:55 PM

how would one solve this?
I used integration by parts and got

(3+2x^2)lnx - 1/2x^2lnx - 1/4x^2

it's a differential equation os

1/3y^3 = (3+2x^2)lnx - 1/2x^2lnx - 1/4x^2 where y(1) = 0

**Isomorphism** · June 6th 2008, 10:00 PM

Originally Posted by khuezy

how would one solve this?
I used integration by parts and got

(3+2x^2)lnx - 1/2x^2lnx - 1/4x^2

it's a differential equation os

1/3y^3 = (3+2x^2)lnx - 1/2x^2lnx - 1/4x^2 where y(1) = 0

By parts is not required, you can do it this way...
$\int \frac{3+2x^2}{x} \, dx = \int \left(\frac{3}{x}+2x\right)\, dx = 3\ln x + x^2 +C$

**mr fantastic** · June 6th 2008, 10:41 PM

Originally Posted by Isomorphism

By parts is not required, you can do it this way...
$\int \frac{3+2x^2}{x} \, dx = \int \left(\frac{3}{x}+2x\right)\, dx = 3\ln x + x^2 +C$

*Ahem* $3\ln {\color{red}|}x{\color{red}|} + x^2 +C$

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