Solveeit Logo
Login

Question

Question: Find the value of A + B, if \( \int{\dfrac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{x}}-4{{e}^{-x}}}dx=Ax+B\ln...

Find the value of A + B, if ∫4ex+6e−x9ex−4e−xdx=Ax+Bln⁡(9e2x−4)+C\int{\dfrac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{x}}-4{{e}^{-x}}}dx=Ax+B\ln \left( 9{{e}^{2x}}-4 \right)+C}
(a) −925-\dfrac{9}{25}
(b) −1936-\dfrac{19}{36}
(c) 1936\dfrac{19}{36}
(d) 935\dfrac{9}{35}

Explanation

Solution

Hint : To solve this question, we will first simplify the given expression as much as possible. Then we will write the numerator as the sum of some factor into denominator and some factor into derivative of denominator. Then we will find the values of those factors and split the numerator and then integrate them. In this way, we can avoid complex integrations. Thus, we will find the integral in the form Ax+Bln⁡(9e2x−4)+CAx+B\ln \left( 9{{e}^{2x}}-4 \right)+C and find the value of A and B. After adding A and B, we get the solution to the question.

Complete step-by-step answer :
To begin with, we will multiply and divide the given expression by ex{{e}^{x}} .
⇒4ex+6e−x9ex−4e−x×exex=4e2x+69e2x−4\Rightarrow \dfrac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{x}}-4{{e}^{-x}}}\times \dfrac{{{e}^{x}}}{{{e}^{x}}}=\dfrac{4{{e}^{2x}}+6}{9{{e}^{2x}}-4}
Now, we need to split the numerator in such a way that one term is multiple of denominator and the other is the multiple of derivative of the denominator.
The denominator with us is 9e2x−49{{e}^{2x}}-4 .
⇒d(9e2x−4)dx=18e2x\Rightarrow \dfrac{d\left( 9{{e}^{2x}}-4 \right)}{dx}=18{{e}^{2x}}
Thus, 4e2x+6=M(9e2x−4)+N(18e2x)4{{e}^{2x}}+6=M\left( 9{{e}^{2x}}-4 \right)+N\left( 18{{e}^{2x}} \right)
Now, we need to find value of M and N.
From the LHS and RHS, we can say that;
9M + 18N = 4
─4M = 6
⇒\Rightarrow M = −32-\dfrac{3}{2}
We will substitute M = −32-\dfrac{3}{2} in 9M + 18N = 4
⇒9×−32+18N=4 ⇒N=3536 \begin{aligned} & \Rightarrow 9\times -\dfrac{3}{2}+18\text{N}=4 \\\ & \Rightarrow \text{N}=\dfrac{35}{36} \\\ \end{aligned}
Thus, 4e2x+6=−32(9e2x−4)+3536(18e2x)4{{e}^{2x}}+6=-\dfrac{3}{2}\left( 9{{e}^{2x}}-4 \right)+\dfrac{35}{36}\left( 18{{e}^{2x}} \right)
Thus, the expression modifies as 4ex+6e−x9ex−4e−x=−32(9e2x−4)+3536(18e2x)9e2x−4\dfrac{4{{e}^{x}}+6{{e}^{-x}}}{9{{e}^{x}}-4{{e}^{-x}}}=\dfrac{-\dfrac{3}{2}\left( 9{{e}^{2x}}-4 \right)+\dfrac{35}{36}\left( 18{{e}^{2x}} \right)}{9{{e}^{2x}}-4}
Thus, we can split the whole expression.
⇒−32(9e2x−4)+3536(18e2x)9e2x−4=−32(1)+3536×18e2x(9e2x−4)\Rightarrow \dfrac{-\dfrac{3}{2}\left( 9{{e}^{2x}}-4 \right)+\dfrac{35}{36}\left( 18{{e}^{2x}} \right)}{9{{e}^{2x}}-4}=-\dfrac{3}{2}\left( 1 \right)+\dfrac{35}{36}\times \dfrac{18{{e}^{2x}}}{\left( 9{{e}^{2x}}-4 \right)}
Now, we shall integrate −32(1)+3536×18e2x(9e2x−4)-\dfrac{3}{2}\left( 1 \right)+\dfrac{35}{36}\times \dfrac{18{{e}^{2x}}}{\left( 9{{e}^{2x}}-4 \right)} .
We shall remember that ∫f′(x)f(x)=ln⁡(f(x))\int{\dfrac{f'\left( x \right)}{f\left( x \right)}=\ln \left( f\left( x \right) \right)}
⇒∫−32(1)+3536×18e2x(9e2x−4)=−32x+3536ln⁡(9e2x−4)+C\Rightarrow \int{-\dfrac{3}{2}\left( 1 \right)+\dfrac{35}{36}\times \dfrac{18{{e}^{2x}}}{\left( 9{{e}^{2x}}-4 \right)}=-\dfrac{3}{2}x+\dfrac{35}{36}\ln \left( 9{{e}^{2x}}-4 \right)+C}
Thus, A = −32-\dfrac{3}{2} and B = 3536\dfrac{35}{36} .
So, A + B = −32+3536-\dfrac{3}{2}+\dfrac{35}{36}
⇒\Rightarrow A + B = −1936-\dfrac{19}{36}
So, the correct answer is “Option B”.

Note : If splitting of the numerator would not have been done, we would have to integrate it by parts. The formula for integration by parts is as follows: ∫uv=u∫v−∫∫v−dudx\int{uv=u\int{v-\int{\int{v-\dfrac{du}{dx}}}}} .