Question #162041

Evaluate the integral of (tan^-1 x)/(1+x² )dx from 0 to1


1
Expert's answer
2021-02-24T07:31:03-0500

Evaluate the integral of (tan^-1 x)/(1+x² )dx from 0 to1


Let u=tan1xu=tan^{-1}x

Then dudx=11+x2{du \over dx}={1 \over 1+x^2}

dx=(1+x2)dudx=(1+x²)du

0u1+x2.(1+x2)du\int_0 {u \over 1+x²}.(1+x²)du


=u22=(tan1x)22{u² \over 2}= {(tan^{-1}x)^2 \over 2} From 0 to 1

(tan11)22(tan10)22{(tan^{-1}1)² \over 2}-{(tan^{-1}0)² \over 2}

(tan11)22{(tan^{-1}1)² \over 2}







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