Q117 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x=$

  • A
    $8+2 \log 2$
  • B
    $8+\log 2$
  • Correct answer
    C
    $8+2 \log 5$
  • D
    $4+\log 5$

Solution

$I=\int_0^{16} \frac{\sqrt{x}}{1+\sqrt{x}} d x=\int_0^{16}\left(1-\frac{1}{1+\sqrt{x}}\right) d x$ $\begin{aligned} & \Rightarrow \text { Put } 1+\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \Rightarrow d x=2 \sqrt{x} d t \\ & \Rightarrow d x=2(t-1) d t \Rightarrow I=[x]_0^{16}-\int_1^5 \frac{2}{t}(t-1) d t \\ & \Rightarrow I=16-2[(t-1 n t)]_1^5=16-2(5-\ln 5-1)=8+2 \ln 5 .\end{aligned}$

Hence, the correct answer is option: 'C'.

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