Question

How do you integrate $\int {{5^{ - x}}dx}$ ?

Answer 1

Hint : Whenever we see a term which cannot be directly integrated through the known formulae, we should first write it in a simpler form and then apply substitution method i.e., substitute a part of the term with a new variable and replace $dx$ with its equivalent in terms of that new variable. After this substitution, we get a form which can directly be integrated through the known formulae.

Complete step-by-step answer :
(i)
We are given,
$\int {{5^{ - x}}dx}$
As we know that negative exponential power means its reciprocal, we can also write it as,
$\int {\dfrac{1}{{{5^x}}}} dx$
(ii)
As this term is not directly integrable, we will substitute ${5^x}$ with $u$ ,
$u = {5^x}$
Taking ${\log _e}$ both sides which can also be written as $\ln$
$\ln u = \ln {5^x}$
By one of the properties of log, we know that:
${\log _a}{x^n} = n{\log _a}x$
Thus, we can write $\ln {5^x}$ as $x\ln 5$
Therefore, we get
$\ln u = x\ln 5$
Differentiating both sides,
$\dfrac{1}{u}du = \ln 5dx$ [because we know that $\dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x}$ ]
Shifting $\ln 5$ to LHS so that we get value of $dx$ in terms of $u$ and $du$ , we get
$dx = \dfrac{{du}}{{(\ln 5)u}}$
(iii)
Now, as we have both ${5^x}$ and $dx$ in terms of $u$ , we can substitute them in $\int {\dfrac{1}{{{5^x}}}} dx$ as
$\int {\dfrac{1}{u}} \times \dfrac{{du}}{{(\ln 5)u}}$
On simplifying, we get:
$\dfrac{1}{{\ln 5}}\int {\dfrac{{du}}{{{u^2}}}}$
(iv)
As we know that,
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}$
We get,
$\dfrac{1}{{\ln 5}}\int {{u^{ - 2}}du = \dfrac{1}{{\ln 5}}(\dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}}) + c}$
On simplifying, we get
$\dfrac{1}{{\ln 5}} \times \dfrac{{{u^{ - 1}}}}{{ - 1}} + c \\\ \- \dfrac{1}{{(\ln 5)u}} + c \\\$ $$
Now, as we know that $u = {5^x}$ , we will write the answer in terms of $x$ again
$- \dfrac{1}{{(\ln 5){5^x}}} + c$
Hence,
$\int {{5^{ - x}}dx = - \dfrac{{{5^{ - x}}}}{{\ln 5}} + c}$
So, the correct answer is “ $\int {{5^{ - x}}dx = - \dfrac{{{5^{ - x}}}}{{\ln 5}} + c}$”.

Note : Always remember to add the constant $c$ after integrating the term as this is one of the most common mistakes a student does. Also, do not replace $dx$ by $du$ directly. You need to calculate it by differentiating $u$ and the term equivalent to $u$ . From there, use the value of $dx$ for substitution.