Question

$\int \sqrt{\frac{x+1}{x}}dx$ is equal to

• A. $\sqrt{x^2+x}+\frac{1}{2}\log |(x+\frac{1}{2})+\sqrt{x^2+x}|+c$
• B. $-\sqrt{x^2+x}+\frac{1}{2}\log |(x+\frac{1}{2})+\sqrt{x^2+x}|+c$
• C. $-\sqrt{x^2+x}+\frac{1}{2}\log |(x+\frac{1}{2})-\sqrt{x^2+x}|$
• D. $\sqrt{x^2+x}+\frac{1}{2}\log |(x+\frac{1}{2})-\sqrt{x^2+x}|$

Answer 1

The correct option is A $\sqrt{x^2+x}+\frac{1}{2}\log |(x+\frac{1}{2})+\sqrt{x^2+x}|+c$

Here, for solving this integral we will try to rationalize the integrand.
Here,
$\sqrt{\frac{x+1}{x}}=\sqrt{\frac{x+1}{x}}\times \sqrt{\frac{x+1}{x+1}}\Rightarrow \sqrt{\frac{x+1}{x}}=\frac{x+1}{\sqrt{x^2+x}}$
Thus, our integral becomes,
$I=\int \frac{(x+1)}{\sqrt{x^2+x}}dx$.
Now, to solve integrals of these form, we express the linear factor of the numerator in terms of derivative of the quadratic factor in the denominator and a constant term.
Thus, we can write:
$(x+1)=a\frac{d}{dx}(x^2+x)+b\Rightarrow (x+1)=a(2x+1)+b\Rightarrow (x+1)=a\left(2x\right)+(a+b)$
Now, comparing the co-efficients of $x$ and constant term, we get
$a=\frac{1}{2},b=\frac{1}{2}$.
Now, our integral becomes:
$I=\frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x}}dx+\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x}}\Rightarrow I=I_1+I_2$
Now, solving for $I_1$
$I_1=\frac{1}{2}\int \frac{2x+1}{\sqrt{x^2+x}}dx\mathrm{Substituting}t=x^2+x,\mathrm{we}\mathrm{get}dt=(2x+1)dx\mathrm{Thus},I_1=\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\sqrt{t}+c{I}_1=\sqrt{x^2+x}+c.$
Similarly $I_2$$I_2=\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x}}\Rightarrow I_2=\frac{1}{2}\int \frac{dx}{\sqrt{x^2+x+\frac{1}{4}-\frac{1}{4}}}\Rightarrow I_2=\frac{1}{2}\int \frac{dx}{\sqrt{{\left(x+\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}\Rightarrow I_2=\frac{1}{2}\log \left(\left(x+\frac{1}{2}\right)+\sqrt{x^2+x}\right|+c$
$\mathrm{Thus},I=I_1+I_2=\sqrt{x^2+x}+\frac{1}{2}\log \left(\left(x+\frac{1}{2}\right)+\sqrt{x^2+x}\right|+c$