1. ## Integration

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Hey every 1 there are two questions here. Btw the integral is over the entire thing in the first question. Ty in advance

2. ## Re: Integration

Originally Posted by JustRandomThings
Question.PNGQuesrions.PNG

Hey every 1 there are two questions here. Btw the integral is over the entire thing in the first question. Ty in advance
$\noindent For the first one, let I = \int \frac{3x^{3} - 2}{x^{2} + 4}\, \mathrm{d}x. Then$

\begin{align*}I &= 3\int \frac{x^{3}}{x^{2} + 4}\, \mathrm{d}x - \int \frac{2}{4 + x^{2}}\, \mathrm{d}x \\ \Rightarrow I &= 3\int \frac{x^{3}}{x^{2} + 4}\, \mathrm{d}x - \tan^{-1} \left(\frac{x}{2}\right). \quad (\star) \end{align*}

$\noindent Note that x^{3} = x\left(x^{2} + 4\right) - 4x \Rightarrow \frac{x^{3}}{x^{2} + 4} = x - \frac{4x}{x^{2} + 4}. So$

\begin{align*}\int \frac{x^{3}}{x^{2} + 4} \, \mathrm{d}x &= \int \left(x - \frac{4x}{x^{2} + 4}\right)\, \mathrm{d}x \\ &= \frac{x^{2}}{2} - 2 \ln \left(x^{2} + 4 \right). \end{align*}

$\noindent Thus from (\star), we have$

\begin{align*} I &= 3\left(\frac{x^{2}}{2} - 2 \ln \left(x^{2} + 4 \right)\right)- \tan^{-1} \left(\frac{x}{2}\right) + c \\ &= \frac{3x^{2}}{2} - 6\ln \left(x^{2} + 4 \right) - \tan^{-1} \left(\frac{x}{2}\right) + c .\end{align*}

3. ## Re: Integration

Thanks! Any ideas for the second question?

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