Question

The integral $\int\dfrac{{dx}}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}{{\left( {x - 3} \right)}^{\left( {\dfrac{6}{7}} \right)}}}}$ is equal to? (c is a constant of integration)
A. $\dfrac{{ - 1}}{{13}}{\left( {\dfrac{{x - 3}}{{x + 4}}} \right)^{ - 13/7}} + c$
B. $\dfrac{1}{2}{\left( {\dfrac{{x - 3}}{{x + 4}}} \right)^{3/7}} + c$
C. $- {\left( {\dfrac{{x - 3}}{{x + 4}}} \right)^{ - 1/7}} + c$
D. ${\left( {\dfrac{{x - 3}}{{x + 4}}} \right)^{1/7}} + c$

Answer 1

Hint : As we can see in the denominator of $\dfrac{1}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}{{\left( {x - 3} \right)}^{\left( {\dfrac{6}{7}} \right)}}}}$ , we have two terms which are multiplied to each other. So we have to write the power of 2nd term in terms of 1st term and consider 1st term as another variable, say t. Now find the derivative of t with respect to x, and put the obtained derivatives in the given integral. Use this info to further solve the question.

Complete step-by-step answer :
We are given to find the value of integral $\int\dfrac{{dx}}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}{{\left( {x - 3} \right)}^{\left( {\dfrac{6}{7}} \right)}}}}$
So first let us consider the denominator of $\dfrac{1}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}{{\left( {x - 3} \right)}^{\left( {\dfrac{6}{7}} \right)}}}}$ , which is ${\left( {x + 4} \right)^{\left( {\dfrac{8}{7}} \right)}}{\left( {x - 3} \right)^{\left( {\dfrac{6}{7}} \right)}}$
The second term of the denominator is ${\left( {x - 3} \right)^{\left( {\dfrac{6}{7}} \right)}}$ , we have to write this term in terms of the power of 1st term.
This means ${\left( {x - 3} \right)^{\left( {\dfrac{6}{7}} \right)}}$ should have a power $\dfrac{8}{7}$ .
${\left( {x - 3} \right)^{\left( {\dfrac{6}{7}} \right)}}$ can also be written as ${\left( {x - 3} \right)^{\left( {\dfrac{{14 - 8}}{7}} \right)}} = {\left( {x - 3} \right)^{2 - \left( {\dfrac{8}{7}} \right)}} = \dfrac{{{{\left( {x - 3} \right)}^2}}}{{{{\left( {x - 3} \right)}^{\left( {\dfrac{8}{7}} \right)}}}}$ , as ${a^{m - n}}$ is also equal to $\dfrac{{{a^m}}}{{{a^n}}}$
Replace the obtained value of 2nd term in the integral $\int\dfrac{{dx}}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}{{\left( {x - 3} \right)}^{\left( {\dfrac{6}{7}} \right)}}}}$
This gives us $\int\dfrac{{dx}}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}\left( {\dfrac{{{{\left( {x - 3} \right)}^2}}}{{{{\left( {x - 3} \right)}^{\left( {\dfrac{8}{7}} \right)}}}}} \right)}}$
$\Rightarrow\int\dfrac{{dx}}{{\left( {\dfrac{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}}}{{{{\left( {x - 3} \right)}^{\left( {\dfrac{8}{7}} \right)}}}}} \right)\left( {{{\left( {x - 3} \right)}^2}} \right)}}$
$\Rightarrow\int\dfrac{{dx}}{{{{\left( {\dfrac{{x + 4}}{{x - 3}}} \right)}^{\left( {\dfrac{8}{7}} \right)}}\left( {{{\left( {x - 3} \right)}^2}} \right)}}$
Let us consider $\dfrac{{x + 4}}{{x - 3}}$ as t.
$\dfrac{{x + 4}}{{x - 3}} = t$
On differentiating t with respect to x, we get
$\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{x + 4}}{{x - 3}}} \right) = \dfrac{{\left( {x - 3} \right)\left( 1 \right) - \left( {x + 4} \right)\left( 1 \right)}}{{{{\left( {x - 3} \right)}^2}}} = \dfrac{{x - 3 - x - 4}}{{{{\left( {x - 3} \right)}^2}}} = \dfrac{{ - 7}}{{{{\left( {x - 3} \right)}^2}}}$
(as $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right)$ is equal to $\dfrac{{v.\left( {\dfrac{{du}}{{dx}}} \right) - u.\left( {\dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$ )
Here u is $\left( {x + 4} \right)$ and v is $\left( {x - 3} \right)$ ; $\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {x + 4} \right) = 1$ and $\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left( {x - 3} \right) = 1$
$\Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{ - 7}}{{{{\left( {x - 3} \right)}^2}}}$
$\Rightarrow \dfrac{{dx}}{{{{\left( {x - 3} \right)}^2}}} = \dfrac{{dt}}{{ - 7}}$
On substituting the values of $\dfrac{{x + 4}}{{x - 3}}$ as t and $\dfrac{{dx}}{{{{\left( {x - 3} \right)}^2}}}$ as $\dfrac{{dt}}{{ - 7}}$ in $\int\dfrac{{dx}}{{{{\left( {\dfrac{{x + 4}}{{x - 3}}} \right)}^{\left( {\dfrac{8}{7}} \right)}}\left( {{{\left( {x - 3} \right)}^2}} \right)}}$ , we get
$\int\dfrac{{dt}}{{ - 7 \times {t^{\left( {\dfrac{8}{7}} \right)}}}}$
$\Rightarrow \dfrac{{ - 1}}{7}\smallint {t^{ - \left( {\dfrac{8}{7}} \right)}}dt$
$\Rightarrow \dfrac{{ - 1}}{7}\left( {\dfrac{{{t^{\left( {\dfrac{{ - 8}}{7}} \right) + 1}}}}{{\left( {\dfrac{{ - 8}}{7}} \right) + 1}}} \right) + c$ , as the integration of ${x^n}$ is $\dfrac{{{x^{n + 1}}}}{{n + 1}}$
$\Rightarrow \dfrac{{ - 1}}{7}\left( {\dfrac{{{t^{\left( {\dfrac{{ - 1}}{7}} \right)}}}}{{\left( {\dfrac{{ - 1}}{7}} \right)}}} \right) + c = \dfrac{{ - 1}}{7}\left( { - 7} \right)\left( {{t^{\left( {\dfrac{{ - 1}}{7}} \right)}}} \right) + c = {t^{\left( {\dfrac{{ - 1}}{7}} \right)}} + c$
On substituting $\dfrac{{x + 4}}{{x - 3}}$ in the place of t, we get
$\Rightarrow {\left( {\dfrac{{x + 4}}{{x - 3}}} \right)^{\left( {\dfrac{{ - 1}}{7}} \right)}} + c$
$\therefore {\left( {\dfrac{{x - 3}}{{x + 4}}} \right)^{\left( {\dfrac{1}{7}} \right)}} + c$ , as ${a^{ - m}} = \dfrac{1}{{{a^m}}}$
The integral $\int\dfrac{{dx}}{{{{\left( {x + 4} \right)}^{\left( {\dfrac{8}{7}} \right)}}{{\left( {x - 3} \right)}^{\left( {\dfrac{6}{7}} \right)}}}}$ is equal to ${\left( {\dfrac{{x - 3}}{{x + 4}}} \right)^{\left( {\dfrac{1}{7}} \right)}} + c$
So, the correct answer is “Option D”.

Note : Do not differentiate when you are asked to integrate. Differentiation is an algebraic expression which is used to determine the change incurred from a point to another whereas Integration is used in calculating the area under the curve, which cannot be easily calculated. The differentiation of sine function is positive cosine, whereas the integration of sine function is negative cosine, as we can see the change is only in the signs. So be careful while integrating or differentiating. Differentiation is the inverse process of integration and vice-versa.