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Question

Mathematics Question on integral

Integrate the function: x−1x2−1\frac {x-1}{\sqrt {x^2-1}}

Answer

∫x−1x2−1 dx∫\frac {x-1}{\sqrt {x^2-1}}\ dx = ∫xx2−1 dx∫\frac {x}{\sqrt {x^2-1}}\ dx - ∫1x2−1 dx∫\frac {1}{\sqrt {x^2-1}}\ dx ……....(1)

For∫xx2−1dx, letx2−1=t⇒2xdx=dtFor ∫\frac {x}{\sqrt {x^2-1}} dx, \ let x^2-1 = t ⇒ 2x dx = dt

∴∫xx2−1dx=12∫dtt∴ ∫\frac {x}{\sqrt {x^2-1}} dx = \frac 12 ∫\frac {dt}{\sqrt t}

=12∫t−12dt=\frac 12 ∫t^{-\frac 12} dt

=12[2t12]=\frac 12[2t^{\frac 12}]

=t=\sqrt t

=x2−1=\sqrt {x^2-1}

From (1),we obtainFrom\ (1), we\ obtain

∫x−1x2−1 dx∫\frac {x-1}{\sqrt {x^2-1}}\ dx = ∫xx2−1 dx∫\frac {x}{\sqrt {x^2-1}}\ dx - ∫1x2−1 dx∫\frac {1}{\sqrt {x^2-1}}\ dx [∫1x2−a2 dt=log ∣x+x2−a2∣][∫\frac {1}{\sqrt {x^2-a^2}} \ dt = log\ |x+\sqrt {x^2-a^2|}]

=x2−1−log∣x+x2−1∣+C= \sqrt {x^2-1}-log|x+\sqrt {x^2-1}|+C