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(dx)/((1+x^(2) )arccotx) What is the equivalent of integral?

Answer ㏑
(1)/(arctanx)+c
Please explain ^-^

User Nobi
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1 Answer

6 votes


\displaystyle\int(\mathrm dx)/((1+x^2)\cot^(-1)x)

Recall that
(\mathrm d)/(\mathrm dt)\cot^(-1)t=-\frac1{1+t^2}. So suppose we substitute
y=\cot^(-1)x, so that
\mathrm dy=-(\mathrm dx)/(1+x^2). Then the integral becomes


\displaystyle-\int\frac{\mathrm dy}y=-\ln|y|+C

and replacing
y for
x gives the antiderivative


-\ln|\cot^(-1)x|+C=\ln\frac1+C

The proposed answer is not correct. Differentiating gives


(\mathrm d)/(\mathrm dx)\ln\frac1{\tan^(-1)x}=-(\mathrm d)/(\mathrm dx)\ln\tan^(-1)x=-((\mathrm d)/(\mathrm dx)\tan^(-1)x)/(\tan^(-1)x)=-\frac1{(1+x^2)\tan^(-1)x}

(Note the negative sign. We can also omit the absolute sign if we use a particular definition for
\cot^(-1)x, but I don't think it's necessary to go into too much detail.)

User Inky
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