Question

Evaluate the integral
${\int }_{-1}^0\frac{1}{(2x^2+4x+7)}dx$

Answer 1

Given : ${\int }_{-1}^0\frac{dx}{({2x}^2+4x+7)}$

$I={\int }_{-1}^0\frac{dx}{({2x}^2+4x+7)}=\frac{1}{2}{\int }_{-1}^0\frac{dx}{(x^2+2x+\frac{7}{2})}=\frac{1}{2}{\int }_{-1}^0\frac{dx}{(x+1{)}^2+{\left(\sqrt{\frac{5}{2}}\right)}^2}=\frac{1}{2\sqrt{\frac{5}{2}}}{\left|{\tan }^{-1}\left[\frac{x+1}{\sqrt{\frac{5}{2}}}\right]\right|}_{-1}^0=\frac{1}{\sqrt{10}}[{\tan }^{-1}\sqrt{\frac{2}{5}}-0]=\frac{1}{\sqrt{10}}{\tan }^{-1}\sqrt{\frac{2}{5}}$

Hence the correct answer is $\frac{1}{\sqrt{10}}{\tan }^{-1}\sqrt{\frac{2}{5}}$