Evaluate-dx-x-2-4x-13-

Question Number 81482 by Bash last updated on 13/Feb/20

E v a l u a t e \int_{- \infty}^{\infty} \frac{d x}{x^{2} + 4 x + 13} .

Commented by abdomathmax last updated on 13/Feb/20

I = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 4 x + 13} = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 4 x + 4 + 9}

= \int_{- \infty}^{+ \infty} \frac{d x}{{(x + 2)}^{2} + 9} =_{x + 2 = 3 u} \int_{- \infty}^{+ \infty} \frac{3 d u}{9 (u^{2} + 1)}

= \frac{1}{3} \int_{- \infty}^{+ \infty} \frac{d u}{u^{2} + 1} = \frac{1}{3} {[a r c t a n u]}_{- \infty}^{+ \infty} = \frac{1}{3} (\frac{π}{2} - (- \frac{π}{2}))

= \frac{π}{3}

Answered by MJS last updated on 13/Feb/20

\int \frac{d x}{x^{2} + 4 x + 13} =

[t = \frac{x + 2}{3} \to d x = 3 d t]

= \frac{1}{3} \int \frac{d t}{t^{2} + 1} = \frac{1}{3} \arctan t = \frac{1}{3} \arctan \frac{x + 2}{3} + C

\underset{- \infty}{\int^{+ \infty}} \frac{d x}{x^{2} + 4 x + 13} = \frac{π}{3}

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