Statement 1: For −1

The discriminant of x2+2(a−1)x+a+5=0is 4(a−1)2−4(a+5)=4a2−3a−4<0 hence the integral ∫dxx2+2(a−1)x+a+5=λtan−1⁡g(x)+CSince the denominator is not zero for any value of x so f is acontinuous function.