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`int1/((x^(2)+1)^(2))`dx=…A. `tan^(-1)x-1/(2x(x^(2)+1))+c`B. `1/2tan^(-1)x+x/(2(x^(2)+1))+c`C. `tan^(-1)x+1/(x^(2)+1)+c`D. `tan^(-1)x+1/(2(x^(2)+1))+c` |
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Answer» Correct Answer - B `int1/(x^(2)+1)dx=1/(x^(2)+1)` `int1dx-int(d/(dx)(1/(x^(2)+1))int1dx)dx` `=x/(x^(2)+1)-int(-2x)/((x^(2)+1)^(2))xdx` (integration by parts) `=x/(x^(2)+1)+2int(x^(2)+1-1)/((x^(2)+1)^(2))dx` `=x/(x^(2)+1)+2intx/((x^(2)+1))dx-2int(dx)/((x^(2)+1)^(2))` .....(i) And also we know that `int(dx)/(x^(2)+1)=tan^(-1)(x)` From Eqs. (i) and (ii), we get `tan^(-1)x=x/(x^(2)+1)+2tan^(-1)x-2int(dx)/((x^(2)+1)^(2))` `rArr2int(dx)/((x^(2)+1)^(2))=tan^(-1)x+x/(x^(2)+1)` `rArrint(dx)/((x^(2)+1)^(2))=1/2tan^(-1)x+x/(2(x^(2)+1))=c` |
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