find-0-1-x-3-3-x-2-x-2-dx-

Question Number 73478 by abdomathmax last updated on 13/Nov/19

f i n d \int_{0}^{1} \frac{x^{3} - 3}{\sqrt{x^{2} - x + 2}} d x

Answered by MJS last updated on 13/Nov/19

\int \frac{x^{3} - 3}{\sqrt{x^{3} - x + 2}} d x = \int \frac{x^{3} - 3}{\sqrt{{(x - \frac{1}{2})}^{2} + \frac{7}{4}}} d x =

[t = \frac{2 x - 1}{\sqrt{7}} \to d x = \frac{\sqrt{7}}{2} d t]

= \frac{1}{8} \int \frac{7 \sqrt{7} t^{3} + 21 t^{2} + 3 \sqrt{7} t - 23}{\sqrt{t^{2} + 1}} d t =

[t = \sinh \ln u = \frac{u^{2} - 1}{2 u} \Rightarrow u = t + \sqrt{t^{2} + 1} \to d t = \frac{\sqrt{t^{2} + 1}}{t + \sqrt{t^{2} + 1}} d u = \frac{u^{2} + 1}{2 u^{2}} d u]

= \frac{1}{64} \int \frac{7 \sqrt{7} u^{6} + 42 u^{5} - 9 \sqrt{7} u^{4} - 268 u^{3} + 9 \sqrt{7} u^{2} + 42 u - 7 \sqrt{7}}{u^{4}} d u

and {it}^{'} s easy to solve this

Commented by abdomathmax last updated on 17/Nov/19

t h a n k y o u s i r .