-0-x dx- 1-x 1-x 2 - -

The correct option is A π/4. Substitute x= tanθ∴ dx= ^2θ dθ . When x= ∞ , tanθ = ∞∴θ = π /2. ∴ I= ∫_0^π /2tanθ ^2θ dθ ( 1+tanθ nbsp; ) ( ...

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Standard XII

Mathematics

Property 1

∫0∞x dx/ 1+x ...

Question

$\int_{0}^{\infty} \frac{x d x}{(1 + x) (1 + x^{2})} = ? ?$

\frac{π}{4} .

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\frac{π}{2} .

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π

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2 π

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Solution

The correct option is A $\frac{π}{4} .$
Substitute $x = tan θ ∴ d x = {sec}^{2} θ d θ .$
When $x = \infty, tan θ = \infty ∴ θ = π / 2.$
$∴ I = \int_{0}^{π / 2} \frac{tan θ {sec}^{2} θ d θ}{(1 + tan θ) ({sec}^{2} θ)} d θ .$
Now change to $sin θ$ and $cos θ .$
$∴ I = \int_{0}^{π / 2} \frac{sin θ d θ}{cos θ + sin θ} = \frac{π}{4}$
$= \int_{0}^{π / 2} \frac{sin (\frac{π}{2} - θ) d θ}{cos (\frac{π}{2} - θ) + sin (\frac{π}{2} - θ)} = \int_{0}^{π / 2} \frac{cos θ d θ}{cos θ + sin θ} = \int_{0}^{\frac{π}{2}} d θ - \int_{0}^{π / 2} \frac{sin θ d θ}{cos θ + sin θ} = \frac{π}{2} - I$
Therefore, $I = \frac{π}{4}$
Ans: A

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