Mathematics

Solve :
$$\displaystyle \int x \sqrt{x^2-1 dx}$$


SOLUTION
$$=\displaystyle \int x\sqrt {x^2 -1}dx$$
Let $$x^2-1=t_1\ 2x\ dx=dt$$
$$\boxed {x\ dx=\dfrac {dt}{2}}$$
$$=\dfrac {1}{2}\displaystyle \int \sqrt {t} dt$$
$$=\dfrac {1}{2}\ \dfrac {t^{3/2}}{3/2}+C$$
$$=\dfrac {1}{2}.\dfrac {2}{3} (x^2-1)^{3/2}+C$$
$$=\dfrac {(x^2 -1)}{3}+C$$
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Subjective Medium Published on 17th 09, 2020
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