Question-14284

Question Number 14284 by tawa tawa last updated on 30/May/17

Answered by Tinkutara last updated on 30/May/17

\int_{\frac{- 3}{2}}^{\infty} \frac{d x}{4 x^{2} + 12 x + 13} = \frac{1}{4} \int_{\frac{- 3}{2}}^{\infty} \frac{d x}{x^{2} + 3 x + \frac{13}{4}}

= \frac{1}{4} \int_{\frac{- 3}{2}}^{\infty} \frac{d x}{{(x + \frac{3}{2})}^{2} + 1^{2}}

= \frac{1}{4} {[\tan^{- 1} (x + \frac{3}{2})]}_{\frac{- 3}{2}}^{\infty}

= \frac{1}{4} [\tan^{- 1} \infty - \tan^{- 1} 0]

= \frac{1}{4} [\frac{π}{2}] = \frac{π}{8}

Commented by tawa tawa last updated on 30/May/17

God bless you sir .

Commented by tawa tawa last updated on 30/May/17

please check the line 3 . is it correct ? or mistake

Commented by tawa tawa last updated on 30/May/17

God bless you sir .