Question

Prove that int dx x^ 2 +a^ 2 = 1 a tan^ -1 x a +C and hence evaluate int dx x^ 2 -6x+13​

Answer 1

Answer:-

See The Attachments for correct explanation and prove...

$\bf \mapsto \int \dfrac{dx}{x {}^{2} - 6x + 13 } = \dfrac{1}{2} { \tan}^{ - 1} \bigg(\dfrac{x - 3}{2} \bigg) + c.$