Question

$\mathrm{If}\int \frac{dx}{\sqrt{x^2-3x+2}}=\log \left(A\right|+C,\mathrm{then}A=$.

• A. $\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+2}$
• B. $\left(x-\frac{3}{2}\right)-\sqrt{x^2-3x+2}$
• C. $\left(x+\frac{3}{2}\right)+\sqrt{x^2-3x+2}$

Answer 1

The correct option is A $\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+2}$

We can solve this integral as:
$I=\int \frac{dx}{\sqrt{x^2-3x+2}}\Rightarrow I=\int \frac{dx}{\sqrt{x^2-2.\frac{3}{2}x+2}}\Rightarrow I=\int \frac{dx}{\sqrt{x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+2}}\Rightarrow I=\int \frac{dx}{\sqrt{{\left(x-\frac{3}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2}}\mathrm{Substituting}t=x-\frac{3}{2},\mathrm{we}\mathrm{get}dt=dx,\mathrm{then},\mathrm{our}\mathrm{integral}\mathrm{becomes}\Rightarrow I=\int \frac{\mathrm{dt}}{\sqrt{t^2-{\left(\frac{1}{2}\right)}^2}}\mathrm{We}\mathrm{know}\mathrm{that}\int \frac{\mathrm{dt}}{\sqrt{t^2-a^2}}=\log \left(t+\sqrt{t^2-a^2}\right|+C,\mathrm{Thus},I=log\left(t+\sqrt{t^2-\frac{1}{4}}\right|+C\mathrm{Now},\mathrm{substituting}\mathrm{back}t\mathrm{we}\mathrm{get},I=log\left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+2}\right|+C$
Thus, comparing we get $A=\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+2}$