let J(x)=∫_0 ^x (t^2 /((√(t+1)) +(√(t+4))))dt find a explicit form of J(x)


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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
$we have J(x)=∫_0 ^x ((t^2 ((√(t+4))−(√(t+1))))/(t+4−t−1)) dt =(1/3) ∫_0 ^x t^2 (√(t+4))dt −(1/3) ∫_0 ^x t^2 (√(t+1))dt changement (√(t+4))=u give t+4 =u^2 ⇒∫_0 ^x t^2 (√(t+4))dt =∫_2^ ^(√(x+4)) (u^2 −4)^2 u (2u)du =2 ∫_2 ^(√(x+4)) u^2 (u^4 −8u^2 +16)du =2 ∫_2 ^(√(x+4)) (u^6 −8u^4 +16u^2 )du =2 [(u^7 /7) −(8/5)u^5 +((16)/3)u^3 ]_2 ^(√(x+4)) =2{(1/7)(x+4)^(7/2) −(8/5)(x+4)^(5/2) +((16)/3)(x+4)^(3/2) −(2^7 /7)+(8/5) 2^5 −((16)/3) 2^3 } also changement (√(t+1))=u give t+1 =u^2 ⇒ ∫_0 ^x t^2 (√(t+1))dt = ∫_1 ^(√(x+1)) (u^2 −1)^2 u (2u)du =2 ∫_1 ^(√(x+1)) u^2 (u^4 −2u^2 +1)du =2 ∫_1 ^(√(x+1)) (u^6 −2u^4 +u^2 )du =2[ (u^7 /7) −(2/5)u^5 +(u^3 /3)]_1 ^(√(x+1)) =2{(1/7)(x+1)^(7/2) −(2/5)(x+1)^(5/2) +(1/3)(x+1)^(3/2) −(1/7) +(2/5) −(1/3)} the value of J(x)is determined..$
$w e h a v e J (x) = \int_{0}^{x} \frac{t^{2} (\sqrt{t + 4} - \sqrt{t + 1})}{t + 4 - t - 1} d t = \frac{1}{3} \int_{0}^{x} t^{2} \sqrt{t + 4} d t - \frac{1}{3} \int_{0}^{x} t^{2} \sqrt{t + 1} d t$ $c h a n g e m e n t \sqrt{t + 4} = u g i v e t + 4 = u^{2} \Rightarrow \int_{0}^{x} t^{2} \sqrt{t + 4} d t = \int_{2^{}}^{\sqrt{x + 4}} {(u^{2} - 4)}^{2} u (2 u) d u$ $= 2 \int_{2}^{\sqrt{x + 4}} u^{2} (u^{4} - 8 u^{2} + 16) d u = 2 \int_{2}^{\sqrt{x + 4}} (u^{6} - 8 u^{4} + 16 u^{2}) d u$ $= 2 {[\frac{u^{7}}{7} - \frac{8}{5} u^{5} + \frac{16}{3} u^{3}]}_{2}^{\sqrt{x + 4}} = 2 {\frac{1}{7} {(x + 4)}^{\frac{7}{2}} - \frac{8}{5} {(x + 4)}^{\frac{5}{2}} + \frac{16}{3} {(x + 4)}^{\frac{3}{2}}$ $- \frac{2^{7}}{7} + \frac{8}{5} 2^{5} - \frac{16}{3} 2^{3}} a l s o c h a n g e m e n t \sqrt{t + 1} = u g i v e t + 1 = u^{2} \Rightarrow$ $\int_{0}^{x} t^{2} \sqrt{t + 1} d t = \int_{1}^{\sqrt{x + 1}} {(u^{2} - 1)}^{2} u (2 u) d u = 2 \int_{1}^{\sqrt{x + 1}} u^{2} (u^{4} - 2 u^{2} + 1) d u$ $= 2 \int_{1}^{\sqrt{x + 1}} (u^{6} - 2 u^{4} + u^{2}) d u = 2 {[\frac{u^{7}}{7} - \frac{2}{5} u^{5} + \frac{u^{3}}{3}]}_{1}^{\sqrt{x + 1}}$ $= 2 {\frac{1}{7} {(x + 1)}^{\frac{7}{2}} - \frac{2}{5} {(x + 1)}^{\frac{5}{2}} + \frac{1}{3} {(x + 1)}^{\frac{3}{2}} - \frac{1}{7} + \frac{2}{5} - \frac{1}{3}}$ $t h e v a l u e o f J (x) i s d e t e r m i n e d . .$
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