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AP and TS EAMCET Engineering Model Paper 20

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Question : 71 of 160

Marks: +1, -0

\int \frac{(x^{2}-1) \, dx}{(x^{2}+1) \sqrt{1+x^{4}}}

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

C

is a real constant then A=

1
2
$\frac{1}{\sqrt{2}}$
$\frac{1}{3}$

Solution:

\int \frac{(x^{2}-1)}{(x^{2}+1) \sqrt{1+x^{4}}} \, dx

= A \tan^{-1} \frac{\sqrt{x^{2}+\frac{1}{x^{2}}}}{\sqrt{2}}+c

I=\int \frac{x^{2}-1}{(x^{2}+1) \sqrt{x^{4}+1}} \, dx

=\int \frac{x^{2} \left(1-\frac{1}{x^{2}}\right)}{x^{2} \left(x+\frac{1}{x}\right) \sqrt{x^{2}+\frac{1}{x^{2}}}} \, dx

Put

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

we have

I=\int \frac{dt}{t \sqrt{t^{2}-2}}

Again putting

t^{2}-2=s^{2}

2 t \, dt=2 s \, ds

I= \int \frac{s \, ds}{(s^{2}+2)s}=\int \frac{ds}{s^{2}+2}

= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{y}{\sqrt{2}} \right)+c

= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{t^{2}-2}}{\sqrt{2}} \right)+c

= \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{x^{2}+\frac{1}{x^{2}}}}{\sqrt{2}} \right)+c

∴

A=\frac{1}{\sqrt{2}}

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