Question

Find $\int {\dfrac{{dx}}{{{x^2} - {a^2}}}}$ and hence evaluate $\int {\dfrac{{dx}}{{{x^2} - 25}}}$.

Answer 1

Integral is either a numerical value equal to the area under the graph of a function, for some interval or a new function the derivative of which is the original function.

Complete step by step solution:
Let I $= \int {\dfrac{1}{{{x^2} - {a^2}}}dx}$
We use the formula,${a^2} - {b^2} = (a - b)(a + b)$
$I = \int {\dfrac{1}{{(x - a)(x + a)}}} dx$
Now, we will multiply numerator and denominator by$2a$, We have
$I = \int {\dfrac{{2a}}{{2a(x - a)(x + a)}}dx}$
$I = \dfrac{1}{{2a}}\int {\dfrac{{2a}}{{(x - a)(x + a)}}} dx$
$I = \dfrac{1}{{2a}}\int {\dfrac{{a + a}}{{(x - a)(x + a)}}} dx$
$I = \dfrac{1}{{2a}}\int {\dfrac{{x - x + a + a}}{{(x - a)(x + a)}}} dx$
$I = \dfrac{1}{{2a}}\int {\dfrac{{x + a - x + a}}{{(x - a)(x + a)}}dx}$
$I = \dfrac{1}{{2a}}\int {\dfrac{{(x + a) - (x - a)}}{{(x - a)(x + a)}}} dx$
I = \dfrac{1}{{2a}}\int {\left\\{ {\dfrac{{x + a}}{{\left( {x + a} \right)\left( {x - a} \right)}} - \dfrac{{x - a}}{{\left( {x + a} \right)\left( {x - a} \right)}}} \right\\}dx}
I = \dfrac{1}{{2a}}\int {\left\\{ {\dfrac{1}{{x - a}} - \dfrac{1}{{x + a}}} \right\\}dx}
$I = \dfrac{1}{{2a}}\left[ {\int {\dfrac{1}{{x - a}}dx} - \int {\dfrac{1}{{x + a}}dx} } \right]$
Using the integral formula $\int {\dfrac{{f'(x)}}{{f(x)}}dx = \log (x) + C}$, we will get
$I = \dfrac{1}{{2a}}[\log \,(x - a) - \log (x + a)] + C$
As we know that log $a - \log b = \log \dfrac{a}{b}$, we have
$I = \dfrac{1}{{2a}}\left[ {\log \left( {\dfrac{{x - a}}{{x + a}}} \right)} \right] + C$
Now, we evaluate$\int {\dfrac{{dx}}{{{x^2} - 25}}}$
We convert $\dfrac{{dx}}{{{x^2} - 25}}$in the form of $\int {\dfrac{{dx}}{{{x^2} - {a^2}}}}$
So, $\int {\dfrac{{dx}}{{{x^2} - {{(5)}^2}}}}$
We will use the result of $\int {\dfrac{{dx}}{{{x^2} - {a^2}}}} = \dfrac{1}{{2a}}\left[ {\log \left( {\dfrac{{x - a}}{{x + a}}} \right)} \right] + C$for solving $\int {\dfrac{{dx}}{{{x^2}{{(5)}^2}}}}$
Therefore, $a = 5$.
Then, $\int {\dfrac{{dx}}{{{x^2} - 25}}} = \dfrac{1}{{2 \times 5}}\left[ {\log \left( {\dfrac{{x - 5}}{{x + 5}}} \right)} \right] + C$
$\int {\dfrac{{dx}}{{{x^2} - 25}} = \dfrac{1}{{10}}\left[ {\log \left( {\dfrac{{x - 5}}{{x + 5}}} \right)} \right] + C}$

Note: In this type of question, students must know that the both integral values have the same sign. Then we will use the previous result for finding the next integral.