Q112 / 160 · AP EAMCET 2004 · Mathematics
$\int \frac{d x}{(x+100) \sqrt{x+99}}=f(x)+c \Rightarrow f(x)$
- A$2(x+100)^{1 / 2}$
- B$3(x+100)^{1 / 2}$
- Correct answerC$2 \tan ^{-1}(\sqrt{x+99})$
- D$2 \tan ^{-1}(\sqrt{x+100})$
Solution
Let
$$
\begin{aligned}
I & =\int \frac{d x}{(x+100) \sqrt{x+99}} \\
& =\int \frac{d x}{\left.(\sqrt{x+99})^2+1\right) \sqrt{x+99}}
\end{aligned}
$$
Put $\sqrt{x+99}=t \Rightarrow \frac{1}{\sqrt{x+99}} d x=2 d t$
$$
\begin{aligned}
\therefore \quad I & =\int \frac{2 d t}{t^2+1}=2 \tan ^{-1} t+c \\
& =2 \tan ^{-1} \sqrt{x+99}+c
\end{aligned}
$$
From Eqs. (i)
$$
\begin{gathered}
2 \tan ^{-1} \sqrt{x+99}+c=f(x)+c \\
f(x)=2 \tan ^{-1} \sqrt{x+99}
\end{gathered}
$$
Hence, the correct answer is option: 'C'.