`int_(0)^(4)(1)/(1+sqrt(x))dx=`. . . .

Options

A. `log((e^(4))/(6))`
B. `log((e^(4))/(3))`
C. `log((e^(4))/(9))`
D. `log((e^(3))/(4))`

Answer

C

Explanation

Let `I=int_(0)^(4)(1)/(1+sqrt(x))dx`
Putting, `1+sqrt(x)=t implies(1)/(2sqrt(x))dx=dt`
`impliesdx=2sqrt(x)dtimpliesdx=2(t-1)dt`
at x=0, t=1 and x=4, t=3
Now, `I=int_(1)^(3)(2(t-1)dt)/(t)=2int_(1)^(3)(1-(1)/(t))dt`
`=2[t-log|t|]_(1)^(3)`
`=2[(3-log3)-(1-log1)]`
`=2[2-log3](becauselog1=0)`
`=4+2log((1)/(3))=4loge+log((1)/(9))`
`=loge^(4)+log((1)/(9))=log((e^(4))/(9))`