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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 71 of 160

Marks: +1, -0

\int\frac{(x-1)dx}{(x+1)\sqrt{x^3+x^2+x}}=A\tan^{-1}\sqrt{f(x)}+

constant, then the ordered pair

(A.f(-1))=

(2, 1)
(2, -1)
(1, 2)
(-2, 2)

Solution:

We have,

\int\;\frac{(x-1) dx}{(x+1) \sqrt{x^{3}+x^{2}+x}} = A \tan^{-1} \sqrt{f(x)}+C

Let

\;\; I=\int\;\frac{(x-1) dx}{(x+1) \sqrt{x^{3}+x^{2}+x}}

=\int\;\frac{x-1}{x(x+1)\sqrt{x+\frac{1}{x}+1}} dx

Put,

x+\frac{1}{x}+1=t

and

\left(1-\frac{1}{x^{2}}\right) dx = dt

=\int\;\frac{(x-1)}{x(x+1)\sqrt{t}} \times \frac{x^{2}}{(x^{2}-1)} dt

=\int\;\frac{x}{(x+1)^{2}\sqrt{t}} dt=\int\;\frac{dt}{\frac{x^{2}+2x+1}{x}\sqrt{t}}

=\int\;\frac{dt}{\left(1+x+\frac{1}{x}+1\right)\sqrt{t}}=\int\;\frac{dt}{(1+t)\sqrt{t}}

=2\tan^{-1}\sqrt{t}+C=2\tan^{-1}\sqrt{x+\frac{1}{x}+1}+C

\therefore A\;\;=2 \Rightarrow f(x)=x+\frac{1}{x}+1

A\;\;=2 \Rightarrow f(-1)=-1-1+l=-1

\therefore \;\; (A, f(-1) \;\; = (2,-1)

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