Familiar with it?

Calculus Level 3

$\displaystyle\int_{2-\ln 3}^{3+\ln 3} \dfrac{\ln (4+x)}{\ln (4+x)+\ln (9-x)} \, \text{dx} = \ ?$

The answer is 1.599.

1 solution

Hasan Kassim
Mar 31, 2015

$\displaystyle I= \int_{2-\log 3}^{3+\log 3} \frac{\log (4+x)}{\log (4+x) +\log (9-x) } dx$

$\displaystyle {\color{#D61F06}{x\to (5-x)}} \; \; \; \; = \int_{2-\log 3}^{3+\log 3} \frac{\log (9-x)}{\log (4+x) +\log (9-x) } dx$

$\displaystyle => 2I = \int_{2-\log 3}^{3+\log 3} \frac{\log (4+x) + \log (9-x) }{\log (4+x) +\log (9-x) } dx$

$\displaystyle = \int_{2-\log 3}^{3+\log 3} dx = (3+\log 3) - (2-\log 3) = 1+2\log 3$

$\displaystyle => \boxed{I= \frac{1+2\log 3}{2} }$

Moderator note:

Great substitution. But you didn't show that the integrand is continuous and defined in the interval $(2 - \log 3, 3 + \log 3 )$ , if its not fulfilled, you cannot perform such substitution.

Familiar with it?

The answer is 1.599.

1 solution

Moderator note:

0 pending reports