If intleft(frac{4 e^x-25}{2 e^x-5}right) d x= | vedclass

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Question

(B) We are given the integral $I = \int \frac{4 e^x-25}{2 e^x-5} dx$. To solve this,we rewrite the numerator in terms of the denominator: $4 e^x - 25

Vedclass · Accepted Answer

$A=5, B=-3$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{4 e^x-25}{2 e^x-5} dx$. To solve this,we rewrite the numerator in terms of the denominator: $4 e^x - 25 = 2(2 e^x - 5) - 15$. Wait,let's adjust the split: $4 e^x - 25 = 2(2 e^x - 5) - 15$. Actually,a better split is: $4 e^x - 25 = 2(2 e^x - 5) - 15$. Let's re-evaluate: $4 e^x - 25 = 2(2 e^x - 5) - 15$. Wait,$2(2 e^x - 5) = 4 e^x - 10$. So,$4 e^x - 25 = (4 e^x - 10) - 15 = 2(2 e^x - 5) - 15$. Thus,$\int \frac{4 e^x-25}{2 e^x-5} dx = \int \left( 2 - \frac{15}{2 e^x-5} \right) dx$. This does not match the form $Ax + B \log(2 e^x - 5)$. Let's try another approach: Multiply numerator and denominator by $e^{-x}$: $\int \frac{4 - 25 e^{-x}}{2 - 5 e^{-x}} dx$. Let $u = 2 - 5 e^{-x}$,then $du = 5 e^{-x} dx$,so $dx = \frac{du}{5 e^{-x}} = \frac{du}{5(2-u)/5} = \frac{du}{2-u}$. This is complex. Let's use the method: $4 e^x - 25 = A(2 e^x - 5) + B(2 e^x)$. $4 e^x - 25 = (2A + 2B) e^x - 5A$. Comparing coefficients: $-5A = -25 \implies A = 5$. $2A + 2B = 4 \implies 10 + 2B = 4 \implies 2B = -6 \implies B = -3$. So,$\int \frac{4 e^x-25}{2 e^x-5} dx = \int \left( 5 - 3 \frac{2 e^x}{2 e^x-5} \right) dx = 5x - 3 \log|2 e^x - 5| + c$. Therefore,$A=5$ and $B=-3$.

If $\int\left(\frac{4 e^x-25}{2 e^x-5}\right) d x=A x+B \log \left(2 e^x-5\right)+c$ (where $c$ is a constant of integration),then:

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