How do you find $\int {x \times \left( {{2^x}} \right)dx} $ ?

Hint: In the given question, we are required to find the value of the integral given to us in the question by using the integration by parts method. So, we consider the given integral as a new variable. Consider $I = \int {x\left( {{2^x}} \right)dx} $.In integration by parts method, we integrate a function which is a product of two functions using a formula: $\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } $.
We decide the order of the functions using the acronym ILATE where $I$ stands for inverse functions, $L$ for logarithmic functions, $A$ for algebraic functions, $T$ for trigonometric functions, and $E$ for exponential functions.

Complete step by step answer:
Using integration by parts method and considering $x$as first function and ${2^x}$ as second function, we get,
\[I = \left[ {x\int {{2^x}dx} } \right] - \int {\left[ {\dfrac{d}{{dx}}(x).\int {{2^x}dx} } \right]} dx\]
Now, we know that the derivative of x with respect to x is $1$. Also, we know that the integral of ${2^x}$with respect to x is \[\dfrac{{{2^x}}}{{\ln 2}}\] .
\[I = \left[ {x \times \left( {\dfrac{{{2^x}}}{{\ln 2}}} \right)} \right] - \int {\left[ {1 \times \left( {\dfrac{{{2^x}}}{{\ln 2}}} \right)} \right]} dx\]

Simplifying the expression, we get,
\[ \Rightarrow I = \left( {\dfrac{{x{2^x}}}{{\ln 2}}} \right) - \dfrac{1}{{\ln 2}}\int {{2^x}} dx\]
Now, we know that the integral of ${2^x}$with respect to x is $\dfrac{{{2^x}}}{{\ln 2}}$ . Hence, substituting the values of the known integrals, we get the value of original integral as:
\[ \Rightarrow I = \left( {\dfrac{{x{2^x}}}{{\ln 2}}} \right) - \dfrac{1}{{\ln 2}}\left( {\dfrac{{{2^x}}}{{\ln 2}}} \right) + c\]
Opening the brackets and simplifying,
\[ \therefore I = \left( {\dfrac{{x{2^x}}}{{\ln 2}}} \right) - \dfrac{{{2^x}}}{{{{\left( {\ln 2} \right)}^2}}} + c\]

Therefore, the value of integral $\int {x \times \left( {{2^x}} \right)dx} $ is \[\left( {\dfrac{{x{2^x}}}{{\ln 2}}} \right) - \dfrac{{{2^x}}}{{{{\left( {\ln 2} \right)}^2}}} + c\], where c is any arbitrary constant.

Note: Integration by parts method can be used to solve integrals of various complex functions involving the product of functions within itself easily. In the given question, the use of integration by parts method once has made the process much easier and organized. However, this might not be the case every time. We may have to apply the integration by parts method multiple times in order to reach the final answer. We chose the first function to be x as it is an algebraic function and the other function ${2^x}$ is an exponential function and ILATE rule prefers the algebraic function as the first function over the exponential function.