Question

$\int \frac{(4x-1)dx}{\sqrt{2x^2-6x+18}}$ is equal to

• A. $-2\sqrt{2x^2-6x+18}-\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$
• B. $-2\sqrt{2x^2-6x+18}+\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$
• C. $2\sqrt{2x^2-6x+18}-\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$
• D. $2\sqrt{2x^2-6x+18}+\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$

Answer 1

The correct option is D $2\sqrt{2x^2-6x+18}+\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$

Solving integrals of these types recquires expressing the numerator of the integrand as the sum of derivative of the denominator and a constant.
So here, we can express the numerator as:
$(4x-1)=a\frac{d}{dx}(2x^2-6x+18)+b$
$\Rightarrow (4x-1)=a(4x-6)+b$
Thus, comparing the coefficient of x and constant term we get
$a=1,b=5$
Now ,we can write the integral as:
$\Rightarrow I=\int \frac{(4x-1)dx}{\sqrt{2x^2-6x+18}}\Rightarrow I=\int \frac{(4x-6)+5}{\sqrt{2x^2-6x+18}}dx\Rightarrow I=\int \frac{(4x-6)}{\sqrt{2x^2-6x+18}}dx+5\int \frac{dx}{\sqrt{2x^2-6x+18}}\Rightarrow I=2\sqrt{2x^2-6x+18}+I_1$
Now, we can evaluate $I_1$ as:
$I_1=5\int \frac{dx}{\sqrt{2x^2-6x+18}}\Rightarrow I_1=\frac{5}{\sqrt{2}}\int \frac{dx}{\sqrt{x^2-3x+9}}\Rightarrow I_1=\frac{5}{\sqrt{2}}\int \frac{dx}{\sqrt{x^2-3x+\frac{9}{4}-\frac{9}{4}+9}}\Rightarrow I_1=\frac{5}{\sqrt{2}}\int \frac{dx}{\sqrt{{\left(x-\frac{3}{2}\right)}^2-\frac{27}{4}}}\Rightarrow I_1=\frac{5}{\sqrt{2}}\int \frac{dx}{\sqrt{{\left(x-\frac{3}{2}\right)}^2-{\left(\frac{\sqrt{27}}{2}\right)}^2}}\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{formula}\int \frac{dx}{\sqrt{x^2-a^2}}=\ln \left(x+\sqrt{x^2-a^2}\right|+C,\mathrm{we}\mathrm{get}\Rightarrow I_1=\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$
Thus the integral is
$I=2\sqrt{2x^2-6x+18}+\frac{5}{\sqrt{2}}\ln \left(\left(x-\frac{3}{2}\right)+\sqrt{x^2-3x+9}\right|+C$
Thus, Option d. is correct.