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Question: How do you integrate \( \int {\dfrac{{5 - {e^x}}}{{{e^{2x}}}}} dx \) ?...

How do you integrate ∫5−exe2xdx\int {\dfrac{{5 - {e^x}}}{{{e^{2x}}}}} dx ?

Explanation

Solution

n this question, we have to calculate integral of exponential function and algebraic function separately and then add both the final answers to get the required result.

Complete step by step solution:
In this question we are given with the function 5−exe2x\dfrac{{5 - {e^x}}}{{{e^{2x}}}} which can further be simplified as
5e2x+exe2x\dfrac{5}{{{e^{2x}}}} + \dfrac{{{e^x}}}{{{e^{2x}}}} =5e−2x+e−x= 5{e^{ - 2x}} + {e^{ - x}}
Here, we can see that the function is of the form f+gf + g where ff and gg are two different functions.
Hence, we can apply the sum rule of integration which is given as follows:
∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx\int {f(x) \pm g(x)dx = \int {f(x)dx} \pm \int {g(x)dx} }
On comparing 5e−2x+e−x5{e^{ - 2x}} + {e^{ - x}} with the sum rule, we can see that f(x)=5e−2xf(x) = 5{e^{ - 2x}} and g(x)=e−xg(x) = {e^{ - x}}
Thus, on substitution we get,
I=∫5e−2xdx+∫e−xdxI = \int {5{e^{ - 2x}}dx + \int {{e^{ - x}}dx} } ......(1)
We will solve the integral ∫5e−2xdx\int {5{e^{ - 2x}}dx} separately by using the u-substitution method.
Let −2x=u- 2x = u , then we get −12u=x- \dfrac{1}{2}u = x
Differentiate both sides of the equation −12u=x- \dfrac{1}{2}u = x with respect to x, we get
−12du=dx- \dfrac{1}{2}du = dx
Therefore, on substitution we have
∫5e−2xdx=−52∫eudu\int {5{e^{ - 2x}}dx} = - \dfrac{5}{2}\int {{e^u}du}
We know that ∫exdx=ex+c\int {{e^x}dx = } {e^x} + c where c is a constant.
Using this, we get
Now, let us substitute −2x=u- 2x = u . Then we get
∫5e−2xdx=−52e−2x+c1\int {5{e^{ - 2x}}dx} = - \dfrac{5}{2}{e^{ - 2x}} + {c_1} ......(2)
Similarly let us integrate g(x)=e−xg(x) = {e^{ - x}}
⇒∫(g)dx=∫e−xdx\Rightarrow \int {(g)dx = \int {{e^{ - x}}dx} }
We know that ∫exdx=ex+c\int {{e^x}dx = } {e^x} + c where c is a constant.
Using this property we have
⇒∫g(x)dx=e−x−1=−e−x+c2\Rightarrow \int {g(x)dx = \dfrac{{{e^{ - x}}}}{{ - 1}} = - } {e^{ - x}} + {c_2} ......(3)
On adding (2) and (3) we have
∫f(x)dx+∫g(x)dx=−52e−2x−e−x+c1+c2\int {f(x)dx + \int {g(x)dx = - \dfrac{5}{2}{e^{ - 2x}} - {e^{ - x}}} } + {c_1} + {c_2}
⇒I=−52e2x−1ex+c\Rightarrow I = - \dfrac{5}{{2{e^{2x}}}} - \dfrac{1}{{{e^x}}} + c
Where c=c1+c2c = {c_1} + {c_2} and is the constant of integration.
On further simplifying we get
I=−[5+2ex2e2x]+cI = - \left[ {\dfrac{{5 + 2{e^x}}}{{2{e^{2x}}}}} \right] + c
Hence, this is our required answer.
So, the correct answer is “ I=−[5+2ex2e2x]+cI = -\left[ {\dfrac{{5 + 2{e^x}}}{{2{e^{2x}}}}} \right] + c ”.

Note : This is a common tendency among students to substitute dudu or dxdx while using the u-substitution method. However, this will lead you to a wrong answer. The operation of integration, up to an additive constant, is the inverse of the operation of differentiation.