let-J-x-0-x-t-2-t-1-t-4-dt-find-a-explicit-form-of-J-x-

Question Number 57420 by Abdo msup. last updated on 03/Apr/19

l e t J (x) = \int_{0}^{x} \frac{t^{2}}{\sqrt{t + 1} + \sqrt{t + 4}} d t

f i n d a e x p l i c i t f o r m o f J (x)

Commented by maxmathsup by imad last updated on 05/Apr/19

w e h a v e J (x) = \int_{0}^{x} \frac{t^{2} (\sqrt{t + 4} - \sqrt{t + 1})}{t + 4 - t - 1} d t = \frac{1}{3} \int_{0}^{x} t^{2} \sqrt{t + 4} d t - \frac{1}{3} \int_{0}^{x} t^{2} \sqrt{t + 1} d t

c h a n g e m e n t \sqrt{t + 4} = u g i v e t + 4 = u^{2} \Rightarrow \int_{0}^{x} t^{2} \sqrt{t + 4} d t = \int_{2^{}}^{\sqrt{x + 4}} {(u^{2} - 4)}^{2} u (2 u) d u

= 2 \int_{2}^{\sqrt{x + 4}} u^{2} (u^{4} - 8 u^{2} + 16) d u = 2 \int_{2}^{\sqrt{x + 4}} (u^{6} - 8 u^{4} + 16 u^{2}) d u

= 2 {[\frac{u^{7}}{7} - \frac{8}{5} u^{5} + \frac{16}{3} u^{3}]}_{2}^{\sqrt{x + 4}} = 2 {\frac{1}{7} {(x + 4)}^{\frac{7}{2}} - \frac{8}{5} {(x + 4)}^{\frac{5}{2}} + \frac{16}{3} {(x + 4)}^{\frac{3}{2}}

- \frac{2^{7}}{7} + \frac{8}{5} 2^{5} - \frac{16}{3} 2^{3}} a l s o c h a n g e m e n t \sqrt{t + 1} = u g i v e t + 1 = u^{2} \Rightarrow

\int_{0}^{x} t^{2} \sqrt{t + 1} d t = \int_{1}^{\sqrt{x + 1}} {(u^{2} - 1)}^{2} u (2 u) d u = 2 \int_{1}^{\sqrt{x + 1}} u^{2} (u^{4} - 2 u^{2} + 1) d u

= 2 \int_{1}^{\sqrt{x + 1}} (u^{6} - 2 u^{4} + u^{2}) d u = 2 {[\frac{u^{7}}{7} - \frac{2}{5} u^{5} + \frac{u^{3}}{3}]}_{1}^{\sqrt{x + 1}}

= 2 {\frac{1}{7} {(x + 1)}^{\frac{7}{2}} - \frac{2}{5} {(x + 1)}^{\frac{5}{2}} + \frac{1}{3} {(x + 1)}^{\frac{3}{2}} - \frac{1}{7} + \frac{2}{5} - \frac{1}{3}}

t h e v a l u e o f J (x) i s d e t e r m i n e d . .