Question

If $\int {\dfrac{{{{\left( {\sqrt x } \right)}^5}}}{{{{\left( {\sqrt x } \right)}^7} + {x^6}}}dx = a\log \left( {\dfrac{{{x^k}}}{{1 + {x^k}}}} \right)} + c$ , then $a$ , and $k$ are
$\left( a \right){\text{ }}\dfrac{2}{5},\dfrac{5}{2}$
$\left( b \right){\text{ }}\dfrac{1}{5},\dfrac{2}{5}$
$\left( c \right){\text{ }}\dfrac{5}{2},\dfrac{1}{2}$
$\left( d \right){\text{ }}\dfrac{2}{5},\dfrac{1}{2}$

Answer 1

For finding the values we will take the LHS side and framing the equation in such a way that it gets integrated and then after the integration on solving the equation we will get the values for $a$ and $k$ . And in this way, we are going to solve this problem.

Complete step-by-step answer:
So we have the integral given as $\int {\dfrac{{{{\left( {\sqrt x } \right)}^5}}}{{{{\left( {\sqrt x } \right)}^7} + {x^6}}}dx = a\log \left( {\dfrac{{{x^k}}}{{1 + {x^k}}}} \right)} + c$
Now taking the LHS side of the integral, we have
$\Rightarrow \int {\dfrac{{{{\left( {\sqrt x } \right)}^5}}}{{{{\left( {\sqrt x } \right)}^7} + {x^6}}}dx}$
Now the above integral can also be written as
$\Rightarrow \int {\dfrac{{{x^{5/2}}}}{{{x^{7/2}} + {x^6}}}dx}$
And dividing the numerator and denominator by ${x^{5/2}}$ , we get
$\Rightarrow \int {\dfrac{{dx}}{{x + {x^{7/2}}}}}$
Now on taking the ${x^{1/2}}$ common, we get
$\Rightarrow \int {\dfrac{{dx}}{{{x^{1/2}}\left( {{x^3} + {x^{1/2}}} \right)}}}$
On substituting ${x^{1/2}} = t$ , we will get
$\Rightarrow \dfrac{1}{{2{x^{1/2}}}}dx = dt$
And on integrating it, we get
$\Rightarrow 2\int {\dfrac{{dt}}{{{t^6} + t}}}$
And it can also be written as
$\Rightarrow \dfrac{{ - 2}}{5}\int {\dfrac{{du}}{u}}$ $\Rightarrow 2\int {\dfrac{{dt}}{{{t^6}\left( {1 + \dfrac{1}{{{t^5}}}} \right)}}}$
So, on taking $1 + \dfrac{1}{{{t^5}}} = u$ , we get
$\Rightarrow \dfrac{{ - 5}}{{{t^6}}}dt = du$
And it can also be written as
$\Rightarrow \dfrac{{dt}}{{{t^6}}} = \dfrac{1}{5}du$
So the integration will become
$\Rightarrow \dfrac{{ - 2}}{5}\int {\dfrac{{du}}{u}}$
After integration, it will be equal to
$\Rightarrow \dfrac{{ - 2}}{5}\ln \left| u \right| + c$
So on substituting the value $1 + \dfrac{1}{{{t^5}}} = u$ , we get
$\Rightarrow \dfrac{{ - 2}}{5}\ln \left| {1 + \dfrac{1}{{{t^5}}}} \right| + c$
Now on taking the LCM of it, we get
$\Rightarrow \dfrac{{ - 2}}{5}\ln \left| {\dfrac{{{t^5} + 1}}{{{t^5}}}} \right| + c$
So substituting the value ${x^{1/2}} = t$ in the above equation, we get
$\Rightarrow \dfrac{{ - 2}}{5}\ln \left| {\dfrac{{{x^{5/2}} + 1}}{{{x^{5/2}}}}} \right| + c$
On taking the negative sign common, we have
$\Rightarrow \dfrac{{ - 2}}{5}\ln \left| {\dfrac{{{x^{5/2}}}}{{1 + {x^{5/2}}}}} \right| + c$
So on comparing it with the RHS side, we get
$\Rightarrow a = \dfrac{2}{5},k = \dfrac{5}{2}$
Hence, the option $\left( a \right)$ is correct.

Note: For solving this type of question we need to know the integration and rule to do it. Since in this question, nothing much has been used. So we can say that this type of question is a concept based so here we just need to morph the integral in such a way that it follows some identity so that we can integrate it.