Question:

Step 1: Substitution Let $x=t^4$. Then $dx = 4t^3 dt$. The limits of integration change as follows: When $x=1$, $t^4=1 \implies t=1$. When $x=81$, $t^4=81 \implies t=3$. Substitute these into the integral: $$ \int_1^{81}\dfrac{dx}{\sqrt{x}+\sqrt[4]{x}} = \int_1^{3}\dfrac{4t^3 dt}{\sqrt{t^4}+\sqrt[4]{t^4}} $$ $$ = \int_1^{3}\dfrac{4t^3 dt}{t^2+t} $$ Factor out $t$ from the denominator: $$ = \int_1^{3}\dfrac{4t^3 dt}{t(t+1)} $$ $$ = 4\int_1^{3}\dfrac{t^2}{t+1}dt $$ Step 2: Polynomial Division and Integration Perform polynomial division on the integrand $\dfrac{t^2}{t+1}$: $$ \dfrac{t^2}{t+1} = \dfrac{t^2-1+1}{t+1} = \dfrac{(t-1)(t+1)+1}{t+1} = t-1+\dfrac{1}{t+1} $$ Now, integrate the expression: $$ 4\int_1^{3}\left(t-1+\dfrac{1}{t+1}\right)dt = 4\left[\dfrac{t^2}{2}-t+\ln|t+1|\right]_1^{3} $$ Step 3: Evaluation of Definite Integral Evaluate the antiderivative at the upper and lower limits of integration: $$ 4\left[\left(\dfrac{3^2}{2}-3+\ln|3+1|\right) - \left(\dfrac{1^2}{2}-1+\ln|1+1|\right)\right] $$ $$ = 4\left[\left(\dfrac{9}{2}-3+\ln4\right) - \left(\dfrac{1}{2}-1+\ln2\right)\right] $$ $$ = 4\left[\left(\dfrac{9}{2}-\dfrac{6}{2}+\ln4\right) - \left(\dfrac{1}{2}-\dfrac{2}{2}+\ln2\right)\right] $$ $$ = 4\left[\left(\dfrac{3}{2}+\ln4\right) - \left(-\dfrac{1}{2}+\ln2\right)\right] $$ $$ = 4\left[\dfrac{3}{2}+\ln4 + \dfrac{1}{2}-\ln2\right] $$ Combine the constant terms and use logarithm properties ($\ln a - \ln b = \ln(a/b)$): $$ = 4\left[\dfrac{4}{2} + \ln\left(\dfrac{4}{2}\right)\right] $$ $$ = 4\left[2 + \ln2\right] $$ $$ = 8+4\ln2 $$