Ive totally forgot how to integrate and need help with integrating 3x^2-2 / x with steps shown please my exams on friday and im brickin it
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$\displaystyle \int {\frac{{3x^2 - 2}} {x}\,dx} = \int {3x\,dx} - 2\int {\frac{1} {x}\,dx} .$ Does that make sense?
Originally Posted by Krizalid $\displaystyle \int {\frac{{3x^2 - 2}} {x}\,dx} = \int {3x\,dx} - 2\int {\frac{1} {x}\,dx} .$ Does that make sense? some does, but i dont understand where the 2 over x goes?
Originally Posted by jackman some does, but i dont understand where the 2 over x goes? did you mean $\displaystyle \int \frac {3x^2 - 2}x~dx$ or $\displaystyle \int \left( 3x^2 - \frac 2x \right)~dx$ ? Krizalid changed $\displaystyle \int \frac 2x~dx$ to $\displaystyle 2 \int \frac 1x~dx$. which is valid. you should know this
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