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.