How you calculate this? int_0^1sqrtx/((x+3)sqrt(x+3))∫10√x(x+3)√x+3, Using substitution of sqrt(x/(x+3))=t√xx+3=t.
1 Answer
Explanation:
Establishing a couple identities first from
t^2=x/(x+3)" "" "=>" "" "(1/(x+3)=t^2/x" "color(blue)((star)))t2=xx+3 ⇒ (1x+3=t2x (⋆))
1/t^2=(x+3)/x=1+3/x1t2=x+3x=1+3x
3/x=1/t^2-1=(1-t^2)/t^23x=1t2−1=1−t2t2
x/3=t^2/(1-t^2)x3=t21−t2
x=(3t^2)/(1-t^2)" "color(red)((star))x=3t21−t2 (⋆)
dx=(6t(1-t^2)-3t^2(-2t))/(1-t^2)^2dt=(6t)/(1-t^2)^2dt" "color(green)((star))dx=6t(1−t2)−3t2(−2t)(1−t2)2dt=6t(1−t2)2dt (⋆)
Combining
1/(x+3)=t^2/((3t^2)/(1-t^2))=(1-t^2)/3" "color(orange)((star))1x+3=t23t21−t2=1−t23 (⋆)
Then we have the integral:
intsqrtx/((x+3)sqrt(x+3))dx=int1/(x+3)sqrt(x/(x+3))dx∫√x(x+3)√x+3dx=∫1x+3√xx+3dx
=int(1-t^2)/3(t)(6t)/(1-t^2)^2dt=int(6t^2)/(3(1-t^2))dt=∫1−t23(t)6t(1−t2)2dt=∫6t23(1−t2)dt
=2intt^2/(1-t^2)dt=2int(t^2-1+1)/(1-t^2)=2∫t21−t2dt=2∫t2−1+11−t2
=2int(-(1-t^2))/(1-t^2)dt-2int1/(t^2-1)dt=2∫−(1−t2)1−t2dt−2∫1t2−1dt
=-2intdt-2int1/((t+1)(t-1))dt=−2∫dt−2∫1(t+1)(t−1)dt
Partial fraction decomposition on the latter integral gives:
=-2t+int1/(t+1)dt-int1/(t-1)dt=−2t+∫1t+1dt−∫1t−1dt
=-2t+lnabs(t+1)-lnabs(t-1)=−2t+ln|t+1|−ln|t−1|
=-2sqrt(x/(x+3))+lnabs( ( sqrt(x/(x+3))+1) / ( sqrt(x/(x+3))-1) )=−2√xx+3+ln∣∣ ∣ ∣∣√xx+3+1√xx+3−1∣∣ ∣ ∣∣
=-2sqrt(x/(x+3))+lnabs((sqrtx+sqrt(x+3))/(sqrtx-sqrt(x+3)))=−2√xx+3+ln∣∣∣√x+√x+3√x−√x+3∣∣∣
Now applying the bounds:
int_0^1sqrtx/((x+3)sqrt(x+3))dx=(-2sqrt(x/(x+3))+lnabs((sqrtx+sqrt(x+3))/(sqrtx-sqrt(x+3))))|_0^1∫10√x(x+3)√x+3dx=(−2√xx+3+ln∣∣∣√x+√x+3√x−√x+3∣∣∣)∣∣ ∣∣10
=(-2sqrt(1/4)+lnabs((1+sqrt4)/(1-sqrt4)))-(2(0)+lnabs((0+sqrt3)/(0-sqrt3)))=(−2√14+ln∣∣∣1+√41−√4∣∣∣)−(2(0)+ln∣∣∣0+√30−√3∣∣∣)
=-2(1/2)+lnabs(-3)-(0+ln1)=−2(12)+ln|−3|−(0+ln1)
=-1+ln3=−1+ln3
