Question

How you calculate this? `int_0^1sqrtx/((x+3)sqrt(x+3))`, Using substitution of `sqrt(x/(x+3))=t`.

Answer 1

`-1+ln3`

Explanation:

Establishing a couple identities first from `sqrt(x/(x+3))=t`:

`t^2=x/(x+3)" "" "=>" "" "(1/(x+3)=t^2/x" "color(blue)((star)))`

`1/t^2=(x+3)/x=1+3/x`

`3/x=1/t^2-1=(1-t^2)/t^2`

`x/3=t^2/(1-t^2)`

`x=(3t^2)/(1-t^2)" "color(red)((star))`

`dx=(6t(1-t^2)-3t^2(-2t))/(1-t^2)^2dt=(6t)/(1-t^2)^2dt" "color(green)((star))`

Combining `color(blue)((star))` with `color(red)((star))`:

`1/(x+3)=t^2/((3t^2)/(1-t^2))=(1-t^2)/3" "color(orange)((star))`

Then we have the integral:

`intsqrtx/((x+3)sqrt(x+3))dx=int1/(x+3)sqrt(x/(x+3))dx`

`=int(1-t^2)/3(t)(6t)/(1-t^2)^2dt=int(6t^2)/(3(1-t^2))dt`

`=2intt^2/(1-t^2)dt=2int(t^2-1+1)/(1-t^2)`

`=2int(-(1-t^2))/(1-t^2)dt-2int1/(t^2-1)dt`

`=-2intdt-2int1/((t+1)(t-1))dt`

Partial fraction decomposition on the latter integral gives:

`=-2t+int1/(t+1)dt-int1/(t-1)dt`

`=-2t+lnabs(t+1)-lnabs(t-1)`

`=-2sqrt(x/(x+3))+lnabs( ( sqrt(x/(x+3))+1) / ( sqrt(x/(x+3))-1) )`

`=-2sqrt(x/(x+3))+lnabs((sqrtx+sqrt(x+3))/(sqrtx-sqrt(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`

`=(-2sqrt(1/4)+lnabs((1+sqrt4)/(1-sqrt4)))-(2(0)+lnabs((0+sqrt3)/(0-sqrt3)))`

`=-2(1/2)+lnabs(-3)-(0+ln1)`

`=-1+ln3`