Question

If $\int {\dfrac{{{{\left( {\sqrt x } \right)}^5}}}{{{{\left( {\sqrt x } \right)}^7} + {x^6}}}dx} = \lambda \ln \left( {\dfrac{{{x^a}}}{{1 + {x^a}}}} \right) + c$ then $a + \lambda$ is
a. $= 2$
b. $< 2$
c. $> 2$
d. $= 1$

Answer 1

Hint : Here in this question we have to simplify the given function and then we have to integrate the given function. Hence, we obtain the solution, then we have to compare the RHS of the question and obtain a solution then we will obtain the unknown parameter $a$ and $\lambda$ .

Complete step-by-step answer :
Consider the given function $I = \int {\dfrac{{{{\left( {\sqrt x } \right)}^5}}}{{{{\left( {\sqrt x } \right)}^7} + {x^6}}}dx}$ . The square root can be written in the terms of power so we can rewrite the given function as,
$\Rightarrow \int {\dfrac{{{x^{\dfrac{5}{2}}}}}{{{x^{\dfrac{7}{2}}} + {x^6}}}dx}$
Divide both numerator and denominator by ${x^{\dfrac{5}{2}}}$ , so we have
$\Rightarrow \int {\dfrac{{\dfrac{{{x^{\dfrac{5}{2}}}}}{{{x^{\dfrac{5}{2}}}}}}}{{\dfrac{{{x^{\dfrac{7}{2}}} + {x^6}}}{{{x^{\dfrac{5}{2}}}}}}}dx}$
$\Rightarrow \int {\dfrac{{\dfrac{{{x^{\dfrac{5}{2}}}}}{{{x^{\dfrac{5}{2}}}}}}}{{\dfrac{{{x^{\dfrac{7}{2}}}}}{{{x^{\dfrac{5}{2}}}}} + \dfrac{{{x^6}}}{{{x^{\dfrac{5}{2}}}}}}}dx}$
On further simplification
$\Rightarrow \int {\dfrac{{{x^{\dfrac{5}{2} - \dfrac{5}{2}}}}}{{{x^{\dfrac{7}{2} - \dfrac{5}{2}}} + {x^{6 - \dfrac{5}{2}}}}}dx}$
$\Rightarrow \int {\dfrac{{{x^0}}}{{{x^{\dfrac{2}{2}}} + {x^{\dfrac{7}{2}}}}}dx}$
We know that any element power of zero is one
$\Rightarrow \int {\dfrac{{dx}}{{x + {x^{\dfrac{7}{2}}}}}}$
Taking ${x^{\dfrac{1}{2}}}$ as common in the denominator
$\Rightarrow \int {\dfrac{{dx}}{{{x^{\dfrac{1}{2}}}({x^{\dfrac{1}{2}}} + {x^3})}}}$
Substitute ${x^{\dfrac{1}{2}}} = t$
On differentiating the above function w.r.t, x we have
$\dfrac{d}{{dx}}{x^{\dfrac{1}{2}}} = \dfrac{{dt}}{{dx}}$
$\Rightarrow \dfrac{1}{{2{x^{\dfrac{1}{2}}}}}dx = dt$
$\Rightarrow \dfrac{1}{{{x^{\dfrac{1}{2}}}}}dx = 2dt$ and we have ${x^{\dfrac{1}{2}}} = t$ and ${x^3} = {t^6}$
So I can be written as $\Rightarrow \int {\dfrac{{2dt}}{{(t + {t^6})}}}$
Taking ${t^6}$ as common in the denominator we have
$\Rightarrow 2\int {\dfrac{{dt}}{{{t^6}(1 + \dfrac{1}{{{t^5}}})}}}$
Now substitute $1 + \dfrac{1}{{{t^5}}} = u$
On differentiating the above function w.r.t, t we have
$\Rightarrow \dfrac{d}{{dt}}(1 + \dfrac{1}{{{t^5}}}) = \dfrac{{du}}{{dt}}$
$\Rightarrow \dfrac{d}{{dt}}(1) + \dfrac{d}{{dt}}\left( {\dfrac{1}{{{t^5}}}} \right) = \dfrac{{du}}{{dt}}$

$$\Rightarrow \dfrac{{ - 5}}{{{t^6}}}dt = du \\\ \Rightarrow \dfrac{{dt}}{{{t^6}}} = - \dfrac{{du}}{5} \\\$$

So we have $I = - \dfrac{2}{5}\int {\dfrac{{du}}{u}}$
$\Rightarrow I = - \dfrac{2}{5}\ln (u) + c$
Substituting the value of u we have
$\Rightarrow I = - \dfrac{2}{5}\ln \left( {1 + \dfrac{1}{{{t^5}}}} \right) + c$
And substituting the value of t and simplifying we have,
$\Rightarrow I = - \dfrac{2}{5}\ln \left( {1 + \dfrac{1}{{{x^{\dfrac{1}{2}}}}}} \right) + c$
$\Rightarrow I = - \dfrac{2}{5}\ln \left( {\dfrac{{{x^{\dfrac{1}{2}}} + 1}}{{{x^{\dfrac{1}{2}}}}}} \right) + c$
From the question we have $\int {\dfrac{{{{\left( {\sqrt x } \right)}^5}}}{{{{\left( {\sqrt x } \right)}^7} + {x^6}}}dx} = \lambda \ln \left( {\dfrac{{{x^a}}}{{1 + {x^a}}}} \right) + c$ and we got the solution $\Rightarrow I = - \dfrac{2}{5}\ln \left( {\dfrac{{{x^{\dfrac{1}{2}}} + 1}}{{{x^{\dfrac{1}{2}}}}}} \right) + c$ by comparing these two we can rewrite this as
$\lambda \ln \left( {\dfrac{{{x^a}}}{{1 + {x^a}}}} \right) + c = - \dfrac{2}{5}\ln \left( {\dfrac{{{x^{\dfrac{1}{2}}} + 1}}{{{x^{\dfrac{1}{2}}}}}} \right) + c$
Therefore we have the value for $\lambda$ and $a$ that is $\lambda = - \dfrac{2}{5}$ and $a = \dfrac{1}{2}$
Now we have to find the value of $a + \lambda$
By adding we obtain the solution as $\dfrac{1}{2} + \left( { - \dfrac{2}{5}} \right)$
$\Rightarrow \dfrac{1}{2} - \dfrac{2}{5}$
On simplification we have
$\Rightarrow \dfrac{{5 - 2}}{{10}} = \dfrac{3}{{10}}$
Hence the solution will be less than 2 therefore the option b is the correct one.
So, the correct answer is “Option B”.

Note : By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So simplification is needed. We must know the differentiation formulas. The law of exponents or indices are applied to solve this problem.