Question

The value of $\underset{\ln 3}{\overset{\ln 27}{\int }}\frac{x^{1/4}}{x^{1/4}+(4\ln 3-x{)}^{1/4}}dx$ is

• A. $\ln 3$
• B. $2\ln 3$
• C. $3\ln 3$
• D. $0$

Answer 1

The correct option is A $\ln 3$

Given :
$I=\underset{\ln 3}{\overset{\ln 27}{\int }}\frac{x^{1/4}}{x^{1/4}+(4\ln 3-x{)}^{1/4}}dx$
$\Rightarrow =\underset{\ln 3}{\overset{\ln 27}{\int }}\frac{x^{1/4}}{x^{1/4}+(\ln 81-x{)}^{1/4}}dx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{\ln 3}{\overset{\ln 27}{\int }}\frac{(\ln 81-x{)}^{1/4}}{(\ln 81-x{)}^{1/4}+(\ln x{)}^{1/4}}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right),$ we have
$2I=\underset{\ln 3}{\overset{\ln 27}{\int }}1dx$
$\Rightarrow 2I=\ln 27-\ln 3\therefore I=\ln 3$