Answer
$\dfrac{1.1}{3 \ln (2.1)}$
Work Step by Step
Given: $I=\int_{0}^{1} x^2 (2.1)^{x^2} \ dx$
Let us consider that $u=x^3 \implies dx=\dfrac{du}{3x^2}$
Now, we have $I= \int_{0}^{1} x^2 (2.1)^{u} (\dfrac{du}{3x^2})$
or, $=-\dfrac{1}{3} \int_{0}^{1} (2.1)^{u} \ du$
or, $=[\dfrac{(2.1)^{x^3}}{3 \ln 2.1}]_{0}^{1} +C$
or, $=[\dfrac{(2.1)^{1^3}}{3 \ln 2.1}]-[\dfrac{(2.1)^{0^3}}{3 \ln 2.1}]$
or, $=\dfrac{1.1}{3 \ln (2.1)}$
