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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.4
Integrate the functions in Exercises 86 to 108.
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Question : 92 of 271
Marks: +1, -0
racx−1x2−1rac{x-1}{\sqrt{x^2-1}}
Solution:  
Let I=∫x−1x2−1 dxI = \int \frac{x-1}{\sqrt{x^2-1}} \, dx
=∫xx2−1 dx−∫1x2−1 dx= \int \frac{x}{\sqrt{x^2-1}} \, dx - \int \frac{1}{\sqrt{x^2-1}} \, dx =I1−I2 ext(say)= I_1 - I_2 \, ext{(say)}
Now, I1=∫xx2−1 dxI_1 = \int \frac{x}{\sqrt{x^2-1}} \, dx
Put x2−1=tx^2 - 1 = t ⇒2x dx=dt\Rightarrow 2x \, dx = dt
∴I1=12∫dtt\therefore I_1 = \frac{1}{2} \int \frac{dt}{\sqrt{t}} =12∫t−1/2 dt= \frac{1}{2} \int t^{-1/2} \, dt =12×t1/212+C1= \frac{1}{2} \times \frac{t^{1/2}}{\frac{1}{2}} + C_1 =t+C1= \sqrt{t} + C_1 =x2−1+C1= \sqrt{x^2-1} + C_1
and I2=∫1x2−1 dxI_2 = \int \frac{1}{\sqrt{x^2-1}} \, dx =log⁡∣x+x2−1∣+C2= \log \left| x + \sqrt{x^2-1} \right| + C_2
Therefore I=x2−1−log⁡∣x+x2−1∣+C\text{Therefore } I = \sqrt{x^2-1} - \log \left| x + \sqrt{x^2-1} \right| + C
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