Question

Find the value of \(\int \sqrt{4x^2+4x+5} dx\).

(a) \(2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{{(x+\frac{1}{2})^2+1)}}\right ]+ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]\)

(b) \(2\left [\frac{1}{2} \sqrt{(x+\frac{1}{2})^2+1)}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]\)

(c) \(2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{(x+\frac{1}{2})^2+1)}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]\)

(d) \(2\left [(x+\frac{1}{2}) \sqrt{{(x+\frac{1}{2})^2+1)}}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]\)

The question was posed to me by my college professor while I was bunking the class.

This interesting question is from Improper Integrals in chapter Integral Calculus of Engineering Mathematics

Answer 1

Right option is (c) \(2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{(x+\frac{1}{2})^2+1)}\right ]+\frac{1}{2} ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+\frac{1}{2})^2+1} \right ]\)

Easy explanation: Add constant automatically

Given, \(\int \sqrt{4x^2+4x+5} dx=\int 2\sqrt{(x+\frac{1}{2})^2+1^2} dx\)

=\(\int 2\sqrt{t^2+1^2} dt=2\left [\frac{1}{2} t\sqrt{t^2+1}\right ]+\frac{1}{2} ln⁡[t+\sqrt{t^2+1}]\)

=\(2\left [\frac{1}{2} (x+\frac{1}{2}) \sqrt{(x+\frac{1}{2})^2+1)} \right ]+\frac{1}{2}  ln⁡\left [(x+\frac{1}{2})+\sqrt{(x+1/2)^2+1}\right ]\)