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# Repeated application of integration by parts, gives us the reduction formula, if the integrand is dependent on a natural number n.If $\int {\dfrac{{dx}}{{{x^n}\sqrt {ax + b} }}} = - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}} - A\int {\dfrac{{dx}}{{{x^{n - 1}}\sqrt {ax + b} }} + C}$, then A is equal toA) $\dfrac{{2n - 3}}{{2n - 1}}$B) $\dfrac{{2n - 3}}{{2n - 2}}\dfrac{a}{b}$C) $\dfrac{{n - 2}}{{n - 1}}\dfrac{a}{b}$D) $\dfrac{{n - 3}}{{n - 2}}\dfrac{a}{b}$

Last updated date: 20th Jun 2024
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Hint: We will first write the given expression and then bring the integral part with A on the LHS with the other integral part and combine them. After that, we will differentiate the expression we just wrote and thus by simplifying, we will get the answer.

We are given that $\int {\dfrac{{dx}}{{{x^n}\sqrt {ax + b} }}} = - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}} - A\int {\dfrac{{dx}}{{{x^{n - 1}}\sqrt {ax + b} }} + C}$.
Now, we will bring the integral part on the left with the other integral part. Then, we will have with us the following expression:-
$\Rightarrow \int {\dfrac{{dx}}{{{x^n}\sqrt {ax + b} }}} + A\int {\dfrac{{dx}}{{{x^{n - 1}}\sqrt {ax + b} }}} = - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}} + C$
Now, combining the both integrals on the left hand side, we will get the following expression with us:-
$\Rightarrow \int {\left( {\dfrac{1}{{{x^n}\sqrt {ax + b} }} + \dfrac{A}{{{x^{n - 1}}\sqrt {ax + b} }}} \right)dx} = - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}} + C$
Taking the LCM on the left hand side to write the above expression as follows:-
$\Rightarrow \int {\left( {\dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }}} \right)dx} = - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}} + C$
Now, let us differentiate both the sides in the above expression to get the following expression:-
$\Rightarrow \dfrac{d}{{dx}}\left[ {\int {\left( {\dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }}} \right)dx} } \right] = \dfrac{d}{{dx}}\left( { - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}}} \right) + \dfrac{d}{{dx}}\left( C \right)$
Since, we know that differentiation of any constant is zero. Therefore, we get:-
$\Rightarrow \dfrac{d}{{dx}}\left[ {\int {\left( {\dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }}} \right)dx} } \right] = \dfrac{d}{{dx}}\left( { - \dfrac{{\sqrt {ax + b} }}{{(n - 1)b{x^{n - 1}}}}} \right)$
We can write it as:-
$\Rightarrow \dfrac{d}{{dx}}\left[ {\int {\left( {\dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }}} \right)dx} } \right] = \dfrac{1}{{(n - 1)b}}\dfrac{d}{{dx}}\left( { - \dfrac{{\sqrt {ax + b} }}{{{x^{n - 1}}}}} \right)$ ……………(1)
We know that integration and differentiation will cancel out each other and we will use the formula: $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}(u) - u\dfrac{d}{{dx}}(v)}}{{{v^2}}}$.
Applying both of these things in (1), we will get:-
$\Rightarrow \dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }} = \dfrac{1}{{(n - 1)b}}\left[ {\dfrac{{{x^{n - 1}}\dfrac{d}{{dx}}\left( {\sqrt {ax + b} } \right) - \sqrt {ax + b} \dfrac{d}{{dx}}\left( {{x^{n - 1}}} \right)}}{{{{\left( {{x^{n - 1}}} \right)}^2}}}} \right]$ …………..(2)
We will now use the formulas:- $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}{\left( {cx + d} \right)^n} = c{\left( {cx + d} \right)^{n - 1}} \times \dfrac{d}{{dx}}\left( {cx + d} \right) = {c^2}{\left( {cx + d} \right)^{n - 1}}$ (Because $\dfrac{d}{{dx}}(cx + d) = c$ )
Applying these in (2), we will get:-
$\Rightarrow \dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }} = \dfrac{1}{{(n - 1)b}}\left[ {\dfrac{{{x^{n - 1}}\dfrac{a}{{2\sqrt {ax + b} }} - \sqrt {ax + b} \left( {n - 1} \right){x^{n - 2}}}}{{{x^{2n - 2}}}}} \right]$
We can rewrite this as:-
$\Rightarrow \dfrac{{1 + Ax}}{{{x^n}\sqrt {ax + b} }} = - \dfrac{{ax - 2\left( {n - 1} \right)(ax + b)}}{{2b(n - 1){x^n}\sqrt {ax + b} }}$
Now, crossing out the common in denominators from both the sides, we will then get:-
$\Rightarrow 1 + Ax = - \dfrac{{ax - 2\left( {n - 1} \right)(ax + b)}}{{2b(n - 1)}}$
Simplifying the RHS, we will get:-
$\Rightarrow 1 + Ax = - \dfrac{{ax - 2a(n - 1)x - 2b(n - 1)}}{{2b(n - 1)}}$
Taking 1 from addition in LHS to subtraction in RHS, we will get:-
$\Rightarrow Ax = - \dfrac{{ax - 2a(n - 1)x - 2b(n - 1)}}{{2b(n - 1)}} - 1$
Simplifying the RHS, we will get:-
$\Rightarrow Ax = \dfrac{{ - ax + 2a(n - 1)x}}{{2b(n - 1)}}$
Taking the x from multiplication in LHS to division in RHS, we will then get:-
$\Rightarrow A = \dfrac{{ - a + 2a(n - 1)}}{{2b(n - 1)}}$
Simplifying the RHS, we will get:-
$\Rightarrow A = \dfrac{{ - a + 2an - 2a}}{{2bn - 2b}}$
Simplifying the RHS further, we will get:-
$\Rightarrow A = \dfrac{{2an - 3a}}{{2bn - 2b}}$
We can write it as:-
$\Rightarrow A = \dfrac{{(2n - 3)a}}{{(2n - 2)b}}$

Hence, the correct option is (B).

Note: The students must note that if they will try to approach this problem by using integration by parts, they would not get comparable terms on LHS and RHS which then will be extremely difficult to evaluate by opening up everything.
The students must note that integration and differentiation are basically the reverse of each other like if we add and subtract any constant, it would not make any difference. Similarly if we integrate and then differentiate or vice versa to any function, we will get the same function only.