Question Number 57420 by Abdo msup. last updated on 03/Apr/19
$${let}\:{J}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:\:\:\frac{{t}^{\mathrm{2}} }{\sqrt{{t}+\mathrm{1}}\:+\sqrt{{t}+\mathrm{4}}}{dt} \\ $$$${find}\:{a}\:{explicit}\:{form}\:{of}\:{J}\left({x}\right) \\ $$
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..](Q57477.png)
$${we}\:{have}\:{J}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{\mathrm{2}} \left(\sqrt{{t}+\mathrm{4}}−\sqrt{{t}+\mathrm{1}}\right)}{{t}+\mathrm{4}−{t}−\mathrm{1}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{2}} \sqrt{{t}+\mathrm{4}}{dt}\:−\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{2}} \sqrt{{t}+\mathrm{1}}{dt} \\ $$$${changement}\:\sqrt{{t}+\mathrm{4}}={u}\:{give}\:{t}+\mathrm{4}\:={u}^{\mathrm{2}} \:\Rightarrow\int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{2}} \sqrt{{t}+\mathrm{4}}{dt}\:=\int_{\mathrm{2}^{} } ^{\sqrt{{x}+\mathrm{4}}} \left({u}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} {u}\:\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{2}\:\int_{\mathrm{2}} ^{\sqrt{{x}+\mathrm{4}}} {u}^{\mathrm{2}} \left({u}^{\mathrm{4}} −\mathrm{8}{u}^{\mathrm{2}} \:+\mathrm{16}\right){du}\:=\mathrm{2}\:\int_{\mathrm{2}} ^{\sqrt{{x}+\mathrm{4}}} \left({u}^{\mathrm{6}} −\mathrm{8}{u}^{\mathrm{4}} \:+\mathrm{16}{u}^{\mathrm{2}} \right){du} \\ $$$$=\mathrm{2}\:\left[\frac{{u}^{\mathrm{7}} }{\mathrm{7}}\:−\frac{\mathrm{8}}{\mathrm{5}}{u}^{\mathrm{5}} \:+\frac{\mathrm{16}}{\mathrm{3}}{u}^{\mathrm{3}} \right]_{\mathrm{2}} ^{\sqrt{{x}+\mathrm{4}}} =\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{7}}\left({x}+\mathrm{4}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} \:−\frac{\mathrm{8}}{\mathrm{5}}\left({x}+\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:+\frac{\mathrm{16}}{\mathrm{3}}\left({x}+\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right. \\ $$$$\left.−\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{7}}+\frac{\mathrm{8}}{\mathrm{5}}\:\mathrm{2}^{\mathrm{5}} \:−\frac{\mathrm{16}}{\mathrm{3}}\:\mathrm{2}^{\mathrm{3}} \right\}\:\:\:{also}\:{changement}\:\sqrt{{t}+\mathrm{1}}={u}\:{give}\:{t}+\mathrm{1}\:={u}^{\mathrm{2}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{x}} {t}^{\mathrm{2}} \sqrt{{t}+\mathrm{1}}{dt}\:=\:\int_{\mathrm{1}} ^{\sqrt{{x}+\mathrm{1}}} \left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} {u}\:\left(\mathrm{2}{u}\right){du}\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{{x}+\mathrm{1}}} {u}^{\mathrm{2}} \left({u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}\right){du} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{{x}+\mathrm{1}}} \left({u}^{\mathrm{6}} \:−\mathrm{2}{u}^{\mathrm{4}} \:+{u}^{\mathrm{2}} \right){du}\:=\mathrm{2}\left[\:\frac{{u}^{\mathrm{7}} }{\mathrm{7}}\:−\frac{\mathrm{2}}{\mathrm{5}}{u}^{\mathrm{5}} \:+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\sqrt{{x}+\mathrm{1}}} \\ $$$$=\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{7}}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} \:−\frac{\mathrm{2}}{\mathrm{5}}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:+\frac{\mathrm{1}}{\mathrm{3}}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{7}}\:+\frac{\mathrm{2}}{\mathrm{5}}\:−\frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$$${the}\:{value}\:{of}\:{J}\left({x}\right){is}\:{determined}.. \\ $$