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Question

Question: Find the value of \(\int {{2^x}{e^x}} dx \) \((A)\dfrac{{{2^x}{e^x}}}{{1 - \ln 2}} + C\) \((B)\d...

Find the value of 2xexdx\int {{2^x}{e^x}} dx
(A)2xex1ln2+C(A)\dfrac{{{2^x}{e^x}}}{{1 - \ln 2}} + C
(B)2xex1+ln2+C(B)\dfrac{{{2^x}{e^x}}}{{1 + \ln 2}} + C
(C)2xex1+ln2+C(C)\dfrac{{{2^x}{e^x}}}{{ - 1 + \ln 2}} + C
(D)2xexln(2e)+C(D)\dfrac{{{2^x}{e^x}}}{{\ln (2e)}} + C

Explanation

Solution

Here, we will solve the given integral by using the formula of integration by parts and doing some calculation we will get the required answer.

Formula used: udv=uvvdu\int {udv = uv} - \int {vdu}

Complete step-by-step solution:
Let I=2xexdxI = \int {{2^x}{e^x}} dx
We consider 2x{2^x} being uu and ex{e^x} to be vv and use the integration by parts formula on II.
I=2xex2x(ln2)exdxI = {2^x}{e^x} - \int {{2^x}(\ln 2){e^x}dx}
Since ln2\ln 2 is a constant value we can take it outside the integral sign therefore, the equation can be re-written as:
\Rightarrow I=2xexln22xexdxI = {2^x}{e^x} - \ln 2\int {{2^x}{e^x}dx}
Since we know that 2xexdx\int {{2^x}{e^x}} dx is equal to II, the equation can be written as:
\Rightarrow I=2xex(ln2)II = {2^x}{e^x} - (\ln 2)I
On sending the similar term across the == sign we get:
\Rightarrow I+(ln2)I=2xexI + (\ln 2)I = {2^x}{e^x}
On taking II common we get:
\Rightarrow (1+ln2)I=2xex(1 + \ln 2)I = {2^x}{e^x}
Now sending the term across the == sign, it will become the denominator, it can be written as:
\Rightarrow I=2xex1+ln2I = \dfrac{{{2^x}{e^x}}}{{1 + \ln 2}}
Therefore,
\Rightarrow 2xexdx=2xex1+ln2+c\int {{2^x}{e^x}} dx = \dfrac{{{2^x}{e^x}}}{{1 + \ln 2}} + c

Therefore, the correct option is (B)(B)

Note: The sum can be done using another method which is the following:
I=2xexdxI = \int {{2^x}{e^x}} dx
Now using the rule of exponent, we can rewrite the equation as:
\Rightarrow I=(2e)xdxI = \int {{{(2e)}^x}} dx
Here we have to formula we get,
\Rightarrow ax=axloga+c\int {{a^x} = \dfrac{{{a^x}}}{{\log a}}} + c
Since the value 2e2e is a constant value therefore aa can be considered as 2e2e.
On integrating we get:
\Rightarrow I=(2e)xln(2e)+cI = \dfrac{{{{(2e)}^x}}}{{\ln (2e)}} + c
Now the denominator can be re-split into the original form using the rule of exponents, it can be written as:
\Rightarrow I=2xexln(2e)+cI = \dfrac{{{2^x}{e^x}}}{{\ln (2e)}} + c, which is the required answer.
Therefore, the correct option using this method is option (D)(D).
An answer in integration is never strictly a single answer; the answer could vary based on the method used to solve it.
All the basic integration and derivative formulas should be memorized by a student to solve these types of questions.
Integration and derivation are opposites of each other example if integration of a term AA is BB, then the derivative of term BB will be AA.