Question

$\mathrm{If}\int \frac{x+8}{x^2+6x+5}dx=a\ln \left(x^2+6x+5\right|+b\ln \left(\frac{x+1}{x+5}\right|+C\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{ab}\mathrm{will}\mathrm{be}$

• A. $\frac{5}{16}$
• B. $\frac{5}{8}$
• C. $\frac{5}{4}$
• D. $\frac{5}{2}$

Answer 1

The correct option is B $\frac{5}{8}$

Solving these types of equations requires expressing numerator in terms of derivative of the denominator of the integrand and a constant value,
i.e. for integrand of the form $\frac{N}{D}$ we write $N=c\frac{d}{dx}\left(D\right)+d,\mathrm{where}c\mathrm{and}d\mathrm{are}\mathrm{constants}$
here, we can write,
$x+8=c\frac{d}{dx}(x^2+6x+5)+d\Rightarrow x+8=c(2x+6)+d\Rightarrow x+8=2cx+(6c+d)$
Now, comparing the coefficients of $x$ and constant value we get,
$c=\frac{1}{2},d=5$
now, we can write the integral as
$I=\int \frac{x+8}{x^2+6x+5}dx\Rightarrow I=\int \frac{\frac{1}{2}(2x+6)+5}{x^2+6x+5}\Rightarrow I=\frac{1}{2}\int \frac{2x+6}{x^2+6x+5}dx+5\int \frac{dx}{x^2+6x+5}\Rightarrow I=\frac{1}{2}\ln \left(x^2+6x+5\right|+5\int \frac{dx}{(x+1)(x+5)}$
Now, we can solve the second integral by integration by partial fraction.
So, we get the integral as
$I=\frac{1}{2}\ln \left(x^2+6x+5\right|+\frac{5}{4}\ln \left(\frac{x+1}{x+5}\right|+C$
Now, comparing we get,
$a=\frac{1}{2},b=\frac{5}{4}$thus $ab=\frac{5}{8}$
Thus, Option b. is correct.