What is the improper integrals 1/(x^2+2x+2) dx from 0 to +infinity ? .

Jun 1, 2018

$\textcolor{b l u e}{{\int}_{0}^{+ \infty} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx} = {\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx} = 3 \frac{\pi}{4}}$

Explanation:

show below:

${\int}_{0}^{+ \infty} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx} = {\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{x}^{2} + 2 x + 2} \cdot \mathrm{dx}$

${\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{x}^{2} + 2 x + 1 + 1} \cdot \mathrm{dx} =$

${\lim}_{b \rightarrow + \infty} {\int}_{0}^{b} \frac{1}{{\left(x + 1\right)}^{2} + 1} \cdot \mathrm{dx} = {\lim}_{b \rightarrow + \infty} {\left[\arctan \left(x + 1\right)\right]}_{0}^{b} =$

${\lim}_{b \rightarrow + \infty} \left[\arctan \left(a + 1\right) + \arctan \left(1\right)\right] = \left[\arctan \left(+ \infty\right) + \frac{\pi}{4}\right]$

$\frac{\pi}{2} + \frac{\pi}{4} = 3 \frac{\pi}{4}$