Evaluate the integrals \int_{\sqrt{3}}^3\frac{dt}{3+t^2}

Dillard 2021-10-20 Answered
Evaluate the integrals
33dt3+t2
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Expert Answer

Alannej
Answered 2021-10-21 Author has 104 answers
We have to evaluate the integral
33dt3+t2
We know
dxx2+a2=1atan1(xa)
Therefore,
dxx2+a2=1atan1(xa)
Therefore,
33=33dtt2+(3)2
=[13tan1(t3]33
=[13π313tan1(1)]
=[13(π3π4]
=13π12
=π123
=3π36
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