InterviewSolution
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InterviewSolution
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`inttan^(-1)xdx=….+C`A. `(1)/(1+x^(2))`B. `x tan^(-1)x+(1)/(2)log|1+x^(2)|`C. `x tan^(-1)x+(1)/(2).(tan^(-1)x)/(1+x^(2))`D. `x tan^(-1)x-(1)/(2)log|1+x^(2)|` |
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Answer» Correct Answer - D Let `l=int tan^(-1)underset("I")(x).underset("II")(1)dx` On integration by parts, we get `tan^(-1)x.x-int(1)/(1+x^(2)).x dx = x tan^(-1)-l_(1)" …(i)"` where, `" "l_(1)=int(x)/(1+x^(2))dx` Put `1+x^(2)=t rArr x dx=(1)/(2)dt` `therefore" "l_(1)=(1)/(2)int (1)/(t)dt=(1)/(2)log t=(1)/(2)log(1+x^(2))` On putting the value of `l_(1)` in Eq. (i), we get `l=x tan^(-1)x-(1)/(2)log(1+x^(2))+C` |
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