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Question: If \[\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}=A}\sqrt{7-6x-{{x}^{2}}}+B{{\sin }^{-1}}\dfrac{\left( x...

If ∫2x+57−6x−x2=A7−6x−x2+Bsin⁡−1(x+3)4+C\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}=A}\sqrt{7-6x-{{x}^{2}}}+B{{\sin }^{-1}}\dfrac{\left( x+3 \right)}{4}+C, where C is the constant of integration then find out the value of the ordered pair A. (A,B)\left( A,B \right)?
A. (−2,−1)\left( -2,-1 \right)
B. (2,−1)\left( 2,-1 \right)
C. (−2,1)\left( -2,1 \right)
D. (2,1)\left( 2,1 \right)

Explanation

Solution

In order to find out the values of ordered pairs (A, B), we need to find out the integration of the given integral. For find the integration first we will rewrite the terms of denominator i.e. 7−6x−x2=16−(x+3)27-6x-{{x}^{2}}=16-{{\left( x+3 \right)}^{2}}. Then we need to rewrite the given integral into the difference of two integral and we will get ∫2x+57−6x−x2=∫(−(−2x−6)7−6x−x2−116−(x+3)2)dx\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}}=\int{\left( \dfrac{-\left( -2x-6 \right)}{\sqrt{7-6x-{{x}^{2}}}}-\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}} \right)}dx. Then taking each integral one by one, using the formulas of integration we will integrate the following. At last combining the two integration and comparing it to the given question. In this way we will get the values of the ordered pair.

Formula used:
Using the rule of integration, if the integral is of the form;
∫dxa2−x2=sin⁡−1xa+C\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\dfrac{x}{a}+C}

Complete step by step solution:
We have given that,
∫2x+57−6x−x2=A7−6x−x2+Bsin⁡−1(x+3)4+C\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}=A}\sqrt{7-6x-{{x}^{2}}}+B{{\sin }^{-1}}\dfrac{\left( x+3 \right)}{4}+C
Let I be the integral.
Thus,
⇒I=∫2x+57−6x−x2\Rightarrow I=\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}}
As we know that,
Taking the denominator i.e. 7−6x−x27-6x-{{x}^{2}}
We can rewrite the denominator as,
7−6x−x2=16−(x+3)27-6x-{{x}^{2}}=16-{{\left( x+3 \right)}^{2}}
As,
Using the identity(a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},
⇒16−(x+3)2=16−(x2+6x+9)=16−x2−6x−9=7−x2−6x\Rightarrow 16-{{\left( x+3 \right)}^{2}}=16-\left( {{x}^{2}}+6x+9 \right)=16-{{x}^{2}}-6x-9=7-{{x}^{2}}-6x
Now substituting the replacing value in the given integral,
We will obtain,
⇒I=∫2x+57−6x−x2=∫(−(−2x−6)7−6x−x2−116−(x+3)2)dx\Rightarrow I=\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}}=\int{\left( \dfrac{-\left( -2x-6 \right)}{\sqrt{7-6x-{{x}^{2}}}}-\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}} \right)}dx
Therefore,
⇒I=−∫−2x−67−6x−x2dx−∫116−(x+3)2dx\Rightarrow I=-\int{\dfrac{-2x-6}{\sqrt{7-6x-{{x}^{2}}}}dx}-\int{\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}}dx}
Now,
Taking the integral −∫−2x−67−6x−x2dx-\int{\dfrac{-2x-6}{\sqrt{7-6x-{{x}^{2}}}}dx}
As we know that,
Let 7−6x−x2=t7-6x-{{x}^{2}}=t
Differentiate it with respect to ‘x’
ddx(7−6x−x2)=ddxt\dfrac{d}{dx}\left( 7-6x-{{x}^{2}} \right)=\dfrac{d}{dx}t
Or
(−2x−6)dx=dt\left( -2x-6 \right)dx=dt
We have,
−∫−2x−67−6x−x2dx-\int{\dfrac{-2x-6}{\sqrt{7-6x-{{x}^{2}}}}dx}
Now,
Substituting the values‘t’ and ‘dt’
We will obtain,
−∫dtt=−∫dtt12=−∫t−12dt=−2t-\int{\dfrac{dt}{\sqrt{t}}=}-\int{\dfrac{dt}{{{t}^{\dfrac{1}{2}}}}=-\int{{{t}^{-\dfrac{1}{2}}}}dt}=-2\sqrt{t}
Undo the substitution and putting the value of ‘t’
⇒−2t=−27−6x−x2\Rightarrow -2\sqrt{t}=-2\sqrt{7-6x-{{x}^{2}}}
Thus,
⇒−∫−2x−67−6x−x2dx=27−6x−x2\Rightarrow -\int{\dfrac{-2x-6}{\sqrt{7-6x-{{x}^{2}}}}dx}=2\sqrt{7-6x-{{x}^{2}}}
Now,
Taking the integral ∫116−(x+3)2dx\int{\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}}dx}
We can rewrite it as,
⇒∫116−(x+3)2dx=∫142−(x+3)2dx\Rightarrow \int{\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}}dx}=\int{\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( x+3 \right)}^{2}}}}dx}
We can observe that,
Using the rule of integration,
It is of form;
∫dxa2−x2=sin⁡−1xa+C\int{\dfrac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}={{\sin }^{-1}}\dfrac{x}{a}+C}
Thus,
Replacing the value of ‘a’ by 4,
⇒∫142−(x+3)2dx=sin⁡−1(x+34)+C\Rightarrow \int{\dfrac{1}{\sqrt{{{4}^{2}}-{{\left( x+3 \right)}^{2}}}}dx}={{\sin }^{-1}}\left( \dfrac{x+3}{4} \right)+C
Therefore,
Combining both the integrals we will get,
⇒I=∫2x+57−6x−x2=∫(−(−2x−6)7−6x−x2−116−(x+3)2)dx=−27−6x−x2−sin⁡−1(x+34)+C\Rightarrow I=\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}}=\int{\left( \dfrac{-\left( -2x-6 \right)}{\sqrt{7-6x-{{x}^{2}}}}-\dfrac{1}{\sqrt{16-{{\left( x+3 \right)}^{2}}}} \right)}dx=-2\sqrt{7-6x-{{x}^{2}}}-{{\sin }^{-1}}\left( \dfrac{x+3}{4} \right)+C
Thus,
∴∫2x+57−6x−x2=−27−6x−x2−sin⁡−1(x+34)+C\therefore \int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}}=-2\sqrt{7-6x-{{x}^{2}}}-{{\sin }^{-1}}\left( \dfrac{x+3}{4} \right)+C
OR
∴∫2x+57−6x−x2=−27−6x−x2+(−1)sin⁡−1(x+34)+C\therefore \int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}}=-2\sqrt{7-6x-{{x}^{2}}}+\left( -1 \right){{\sin }^{-1}}\left( \dfrac{x+3}{4} \right)+C
We have given that,
∫2x+57−6x−x2=A7−6x−x2+Bsin⁡−1(x+3)4+C\int{\dfrac{2x+5}{\sqrt{7-6x-{{x}^{2}}}}=A}\sqrt{7-6x-{{x}^{2}}}+B{{\sin }^{-1}}\dfrac{\left( x+3 \right)}{4}+C
Comparing it to the integration of the given integral;
The ordered pair (A,B)=(−2,−1)\left( A,B \right)=\left( -2,-1 \right)

Hence, the option (A) is the correct answer.

Note: There are other methods for integration also. These are integration by partial fractions and integration by trigonometric substitution. Students need to remember all the methods, as it is useful to know all the methods for integration so that we can choose one for computation according to the convenience and ease of calculation. We should do the calculations explicitly so that we can avoid making errors.