Question

$\int \sqrt{x^2+2x+5}\mathrm{dx}\mathrm{is}\mathrm{equal}\mathrm{to}$

• A. $\frac{x+1}{2}\sqrt{x^2+2x+5}-2\ln \left|(x+1)+\sqrt{x^2+2x+5}\right|+C$
• B. $-\frac{x+1}{2}\sqrt{x^2+2x+5}-2\ln \left|(x+1)+\sqrt{x^2+2x+5}\right|+C$
• C. $\frac{x+1}{2}\sqrt{x^2+2x+5}+2\ln \left|(x+1)+\sqrt{x^2+2x+5}\right|+C$
• D. $-\frac{x+1}{2}\sqrt{x^2+2x+5}-2\ln \left|(x+1)+\sqrt{x^2+2x+5}\right|+C$

Answer 1

The correct option is C $\frac{x+1}{2}\sqrt{x^2+2x+5}+2\ln \left|(x+1)+\sqrt{x^2+2x+5}\right|+C$

To, solve integrals of the form
$I=\int \sqrt{{\mathrm{ax}}^2+\mathrm{bx}+c}\mathrm{dx}$ we express the quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+c$ as perfect squares.
Here, $\sqrt{x^2+2x+5}=\sqrt{x^2+2x+1-1+5}=\sqrt{x^2+2x+1+4}=\sqrt{{(x+1)}^2+2^2}$
Now, applying the formula
$\int \sqrt{x^2+a^2}\mathrm{dx}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln \left(x+\sqrt{x^2+a^2}\right|+C$
and similarly substituting
$t=x+1,\mathrm{we}\mathrm{get}\mathrm{dt}=\mathrm{dx}\mathrm{and}\mathrm{integral}\mathrm{becomes}\int \sqrt{t^2+2^2}\mathrm{dt}=\frac{t}{2}\sqrt{t^2+4}+\frac{4}{2}\ln \left(t+\sqrt{t^2+2^2}\right|+C=\frac{t}{2}\sqrt{t^2+4}+2\ln \left(t+\sqrt{t^2+2^2}\right|+C\mathrm{Now},\mathrm{substituting}\mathrm{back}t=x+1,\mathrm{we}\mathrm{get}I=\frac{x+1}{2}\sqrt{(x+1{)}^2+4}+2\ln \left((x+1)+\sqrt{(x+1{)}^2+2^2}\right|+C\Rightarrow I=\frac{x+1}{2}\sqrt{x^2+2x+5}+2\ln \left((x+1)+\sqrt{x^2+2x+5}\right|+C$