2x-3-dx-2x-2-4x-3-

Question Number 95547 by Fikret last updated on 25/May/20

\int \frac{2 x^{3} d x}{2 x^{2} - 4 x + 3} = ?

Answered by MJS last updated on 25/May/20

\int \frac{2 x^{3}}{2 x^{2} - 4 x + 3} d x = \int (x + 2 + \frac{5 x - 6}{2 x^{2} - 4 x + 3}) d x =

= \int (x + 2) d x + \frac{5}{4} \int \frac{4 x - 4}{2 x^{2} - 4 x + 3} d x - \int \frac{d x}{2 x^{2} - 4 x + 3} =

= \frac{1}{2} x^{2} + 2 x + \frac{5}{4} \ln (2 x^{2} - 4 x + 3) - \frac{\sqrt{2}}{2} \arctan (\sqrt{2} (x - 1)) + C

Commented by peter frank last updated on 26/May/20

thank you

Answered by 1549442205 last updated on 26/May/20

we have \frac{{2 x}^{3}}{{2 x}^{2} - 4 x + 3} = x + 2 + \frac{5 x - 6}{{2 x}^{2} - 4 x + 3}

x + 2 + \frac{5 (x - 1) - 1}{{2 x}^{2} - 4 x + 3}

x + 2 + \frac{\frac{5}{4} (4 x - 4) - 1}{{2 x}^{2} - 4 x + 3} = x + 2 + \frac{\frac{5}{4} (4 x - 4)}{{2 x}^{2} - 4 x + 3}

- \frac{1}{{2 x}^{2} - 4 x + 3} . Hence, denoting I the

integral above we have :

I = \frac{x^{2}}{2} + 2 x + \frac{5}{4} \ln ({2 x}^{2} - 4 x + 3) - \int \frac{dx}{{2 x}^{2} - 4 x + 3}

J = \int \frac{dx}{{2 x}^{2} - 4 x + 3} = \frac{1}{2} \int \frac{dx}{{(x - 1)}^{2} + {(\frac{\sqrt{2}}{2)})}^{2}}

= \sqrt{2} \tan^{- 1} [\sqrt{2} (x - 1)] + C . Thus,

I = \frac{x^{2}}{2} + 2 x + \frac{5}{4} \ln ({2 x}^{2} - 4 x + 3) - \frac{\sqrt{2}}{2} \tan^{- 1} [\sqrt{2} (x - 1)] + C