Integration (1 Viewer)

JustRandomThings · Jan 15, 2017

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

InteGrand · Jan 15, 2017

JustRandomThings said:
View attachment 33708 View attachment 33709

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$

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

JustRandomThings · Jan 15, 2017

Thanks! Any ideas for the second question?

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Integration (1 Viewer)

JustRandomThings

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