Q111 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
If $\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}}=A \cdot \tan ^{-1} \sqrt{f(x)}+$ constant, then the ordered pair $(A, f(-1))=$
- A$(2,1)$
- Correct answerB$(2,-1)$
- C$(1,2)$
- D$(-2,2)$
Solution
We have,
$\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}}=A \tan ^{-1} \sqrt{f(x)}+C$
Let
$\begin{aligned}
I & =\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}} \\
& =\int \frac{x-1}{x(x+1) \sqrt{x+\frac{1}{x}+1}} d x
\end{aligned}$
Put,
$\begin{aligned}
x+\frac{1}{x}+1 & =t \text { and }\left(1-\frac{1}{x^2}\right) d x=d t \\
& =\int \frac{(x-1)}{x(x+1) \sqrt{t}} \times \frac{x^2}{\left(x^2-1\right)} d t \\
& =\int \frac{x}{(x+1)^2 \sqrt{t}} d t=\int \frac{d t}{\frac{x^2+2 x+1}{t}} \\
& =\int \frac{d t}{\left(1+x+\frac{1}{x}+1\right) \sqrt{t}}=\int \frac{d t}{(1+t) \sqrt{t}} \\
= & 2 \tan ^{-1} \sqrt{t}+C=2 \tan ^{-1} \sqrt{x+\frac{1}{x}+1}+C
\end{aligned}$
$\begin{aligned} & \therefore \quad A=2 \Rightarrow f(x)=x+\frac{1}{x}+1 \\ & A=2 \Rightarrow f(-1)=-1-1+1=-1 \\ & \therefore \quad(A, f(-1)=(2,-1) \\ & \end{aligned}$
Hence, the correct answer is option: 'B'.