`int_(0)^(4)1/(1+sqrtx)dx`=…..

Options

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

Answer

C

Explanation

Let `I=int_(0)^(4)1/(1+sqrtx)dx`
Putting, `1+sqrtx=t!1/(2sqrtx)dx=dt`
`!dx=2sqrtxdt!dx=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-logabst]_(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)`