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Let `u=int_0^oo (dx)/(x^4+7x^2+1` and `v=int_0^x (x^2dx)/(x^4+7x^2+1)` thenA. `pi//3`B. `pi//6`C. `pi//12`D. `pi//9` |
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Answer» Correct Answer - B `u=int_(0)^(oo) (dx)/(x^(4)+7x^(2)+1)` and `v=int_(0)^(oo) (x^(2)dx)/(x^(4)+7x^(2)+1)` `:. u+v=int_(0)^(oo) (1+x^(2))/(x^(4)+7x^(2)+1)dx` `=int_(0)^(oo) (1/(x^(2))+1)/((x-1/x)^(2)+9)dx` `=1/3["tan"^(-1)((x-1/x)/3)]_(0)^(oo)` `=1/3[pi//2+pi//2]=pi//3` `:.u+v=pi//3` Now `u-v=int_(0)^(oo) (1-x^(2))/(x^(4)+7x^(2)+1)dx` Let `x=1/t` or `x=-(dt)/(t^(2))` `:.u-v=int_(oo)^(0)(1-1/(t^(2)))/(1/(t^(4))+7/(t^(2))+1)(1-1/(t^(2)))dt` `=-int_(0)^(oo)(1-t^(2))/(t^(4)+7t^(2)+1) dt` `=-(u-v)` `:. u-v=0` From 1 and 2, we get `u=v=pi//6` |
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