If int left( frac{4 e^x - 25}{2 e^x - 5} righ | vedclass

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(B) Let $I = \int \left( \frac{4 e^x - 25}{2 e^x - 5} \right) dx$. We can rewrite the numerator as: $4 e^x - 25 = 2(2 e^x - 5) - 15$. So,$I = \int \le

Vedclass · Accepted Answer

$A = 5, B = -3$

Vedclass · Answer

(B) Let $I = \int \left( \frac{4 e^x - 25}{2 e^x - 5} \right) dx$. We can rewrite the numerator as: $4 e^x - 25 = 2(2 e^x - 5) - 15$. So,$I = \int \left( \frac{2(2 e^x - 5) - 15}{2 e^x - 5} \right) dx$. $I = \int \left( 2 - \frac{15}{2 e^x - 5} \right) dx$. Wait,let us re-evaluate the expression: $I = \int \left( \frac{2(2 e^x - 5) - 15}{2 e^x - 5} \right) dx = \int 2 dx - 15 \int \frac{1}{2 e^x - 5} dx$. Alternatively,using the form $\frac{4 e^x - 25}{2 e^x - 5} = \frac{2(2 e^x - 5) - 15}{2 e^x - 5} = 2 - \frac{15}{2 e^x - 5}$. Let us check the provided solution steps: $I = \int \left( \frac{10 e^x - 25 - 6 e^x}{2 e^x - 5} \right) dx = \int \left( \frac{5(2 e^x - 5) - 6 e^x}{2 e^x - 5} \right) dx = \int 5 dx - 6 \int \frac{e^x}{2 e^x - 5} dx$. Let $u = 2 e^x - 5$,then $du = 2 e^x dx$,so $e^x dx = \frac{du}{2}$. $I = 5x - 6 \int \frac{1}{u} \cdot \frac{du}{2} = 5x - 3 \int \frac{1}{u} du = 5x - 3 \log |2 e^x - 5| + C$. Comparing with $Ax + B \log |2 e^x - 5| + C$,we get $A = 5$ and $B = -3$.

If $\int \left( \frac{4 e^x - 25}{2 e^x - 5} \right) dx = Ax + B \log |2 e^x - 5| + C$,then:

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