# Question #fefc1

Apr 24, 2017

$I = \frac{1}{2} \left(3 a - 2 {a}^{2} - {a}^{4} + 9 \ln | a - 3 \frac{|}{|} {a}^{2} - 3 | + {a}^{2} \ln \left({\left(3 {a}^{2} - {a}^{4}\right)}^{{a}^{2}} / \left(3 a - {a}^{2}\right)\right)\right)$

#### Explanation:

$I = \int u \mathrm{dv} = u v - \int v \mathrm{du}$

$u = \ln \left(3 x - {x}^{2}\right) \implies \mathrm{du} = \frac{3 - 2 x}{3 x - {x}^{2}} \mathrm{dx}$

$\mathrm{dv} = x \mathrm{dx} \implies v = \int x \mathrm{dx} = {x}^{2} / 2$

$I = {x}^{2} / 2 \ln \left(3 x - {x}^{2}\right) - \int {x}^{2} / 2 \frac{3 - 2 x}{3 x - {x}^{2}} \mathrm{dx}$

$I = \frac{1}{2} \left[{x}^{2} \ln \left(3 x - {x}^{2}\right) - \int \frac{{x}^{2} \left(2 x - 3\right)}{{x}^{2} - 3 x} \mathrm{dx}\right]$

${I}_{1} = \int \frac{2 {x}^{3} - 3 {x}^{2}}{{x}^{2} - 3 x} \mathrm{dx} = \int \frac{2 {x}^{3} - 6 {x}^{2} + 3 {x}^{2} - 9 x + 9 x}{{x}^{2} - 3 x} \mathrm{dx}$

${I}_{1} = \int \frac{2 x \left({x}^{2} - 3 x\right) + 3 \left({x}^{2} - 3 x\right) + 9 x}{{x}^{2} - 3 x} \mathrm{dx}$

${I}_{1} = \int \left(2 x + 3 + \frac{9 x}{x \left(x - 3\right)}\right) \mathrm{dx} = \int \left(2 x + 3 + \frac{9}{x - 3}\right) \mathrm{dx}$

${I}_{1} = {x}^{2} + 3 x + 9 \ln | x - 3 | + C$

$I = \frac{1}{2} \left[{x}^{2} \ln \left(3 x - {x}^{2}\right) - {x}^{2} - 3 x - 9 \ln | x - 3 |\right] {|}_{a}^{{a}^{2}}$

$I = \frac{1}{2} \left({a}^{4} \ln \left(3 {a}^{2} - {a}^{4}\right) - {a}^{4} - 3 {a}^{2} - 9 \ln | {a}^{2} - 3 | - {a}^{2} \ln \left(3 a - {a}^{2}\right) + {a}^{2} + 3 a + 9 \ln | a - 3 |\right)$

$I = \frac{1}{2} \left(3 a - 2 {a}^{2} - {a}^{4} + 9 \ln | a - 3 \frac{|}{|} {a}^{2} - 3 | + {a}^{2} \ln \left({\left(3 {a}^{2} - {a}^{4}\right)}^{{a}^{2}} / \left(3 a - {a}^{2}\right)\right)\right)$