-d x-x-1- -x2-1-A- -x-1-x-1B- -x-1-x-1C- -x2-1-x-1D- -x-1-x2-1

The correct option is B √(x 1/x+1)Let x+1 = 1/t⇒ dx = 1/t^2 dt ⇒ I = ∫1/t^2dt/1/t√( ( 1/t 1 )^2 1)⇒ I= ∫dt/√(1 2t)⇒ I=√(1 2t)+c ∴ I = √(x 1/x+1)+c ...

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Byju's Answer

Standard XII

Mathematics

Integration of Irrational Algebraic Fractions - 1

∫d x/x+1] √x2...

Question

$\int \frac{d x}{(x + 1) \sqrt{x^{2} - 1}} =$

\sqrt{\frac{x + 1}{x - 1}}

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\sqrt{\frac{x - 1}{x + 1}}

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\sqrt{\frac{x + 1}{x^{2} + 1}}

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\sqrt{\frac{x^{2} + 1}{x - 1}}

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Solution

The correct option is B $\sqrt{\frac{x - 1}{x + 1}}$
Let $x + 1 = \frac{1}{t} \Rightarrow d x = \frac{- 1}{t^{2}} d t \Rightarrow I = - \int \frac{\frac{1}{t^{2}} d t}{\frac{1}{t} \sqrt{{(\frac{1}{t} - 1)}^{2} - 1}} \Rightarrow I = - \int \frac{d t}{\sqrt{1 - 2 t}} \Rightarrow I = \sqrt{1 - 2 t} + c ∴ I = \sqrt{\frac{x - 1}{x + 1}} + c$

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∫dx(x+1)√x2−1=

The correct option is B √x−1x+1Let x+1=1t⇒dx=−1t2dt⇒I=−∫1t2dt1t√(1t−1)2−1⇒I=−∫dt√1−2t⇒I=√1−2t+c∴I=√x−1x+1+c

$\int \frac{d x}{(x + 1) \sqrt{x^{2} - 1}} =$