Question

Evaluate the integrals: $\int^1_{-1} {\frac{dx}{x^2+2x+5}$

Answer 1

Answer:

Value of the given integral is

$\frac{\pi}{8}$

Step-by-step explanation:

Formula used:

$\int{\frac{1}{x^2+a^2}}\:dx=\frac{1}{a}tan^{-1}(\frac{x}{a})+c$

Now,

$\int\limits^1_{-1}{\frac{1}{x^2+2x+5}}\:dx\\\\=\int\limits^1_{-1}{\frac{1}{(x^2+2x+1)+4}}\:dx\\\\=\int\limits^1_{-1}{\frac{1}{(x+1)^2+2^2}}\:dx\\\\=\frac{1}{2}.[tan^{-1}(\frac{x+1}{2})]^1_{-1}\\\\=\frac{1}{2}.[tan^{-1}(\frac{1+1}{2})-tan^{-1}(\frac{-1+1}{2})]\\\\=\frac{1}{2}.[tan^{-1}(\frac{2}{2})-tan^{-1}(\frac{0}{2})]\\\\=\frac{1}{2}.[tan^{-1}(1)-tan^{-1}(0)]\\\\=\frac{1}{2}.[\frac{\pi}{4}]\\\\=\frac{\pi}{8}$