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In this question you must show all stages of your working. $\newline$ Solutions relying entirely on calculator technology are not acceptable. $\newline$ i) Use the substitution $\newline$ $u=e^{x}-3$ to show that $\newline$ $\int_{\ln 5}^{\ln 7}\frac{4e^{3x}}{e^{x}-3}\,dx=a+b \ln 2$ $\newline$ where $\newline$ $a$ and $\newline$ $b$ are constants to be found.

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Simplify radical expressions with variables

Full solution

Q. In this question you must show all stages of your working.

\newline

Solutions relying entirely on calculator technology are not acceptable.

\newline

i) Use the substitution

\newline

u=e^{x}-3

to show that

\newline

\int_{\ln 5}^{\ln 7}\frac{4e^{3x}}{e^{x}-3}\,dx=a+b \ln 2

\newline

where

\newline

a

and

\newline

b

are constants to be found.

Substitution of $u$ : Let's start by making the substitution $u = e^x - 3$ . We need to express everything in the integral in terms of $u$ , including the differential $dx$ and the limits of integration.
Finding $\frac{du}{dx}$ : First, we find $\frac{du}{dx}$ . Since $u = e^x - 3$ , we have $\frac{du}{dx} = e^x$ . Therefore, $du = e^x dx$ .
Changing Limits of Integration: Now we need to change the limits of integration. When $x = \ln(5)$ , $u = e^{\ln(5)} - 3 = 5 - 3 = 2$ . When $x = \ln(7)$ , $u = e^{\ln(7)} - 3 = 7 - 3 = 4$ . So the new limits of integration are from $u = 2$ to $u = 4$ .
Rewriting Integral in terms of u: We can now rewrite the integral in terms of u. The integral becomes $\int \frac{4e^{3x}}{e^{x}-3}dx = \int \frac{4e^{2x} \cdot e^x}{u} \cdot \left(\frac{du}{e^x}\right) = \int \frac{4e^{2x}}{u} du$ .
Expressing $e^{2x}$ in terms of $u$ : We need to express $e^{2x}$ in terms of $u$ . Since $u = e^x - 3$ , we have $e^x = u + 3$ . Therefore, $e^{2x} = (u + 3)^2$ .
Substituting $e^{2x}$ in Integral: Substitute $e^{2x}$ with $(u + 3)^2$ in the integral. The integral now becomes $\int\frac{4(u + 3)^2}{u}\, du$ .
Expanding the Integrand: Expand the integrand: $4(u + 3)^2/u = 4(u^2 + 6u + 9)/u = 4u + 24 + 36/u$ .
Integrating Term by Term: Now we integrate term by term from $u = 2$ to $u = 4$ : $\int(4u + 24 + \frac{36}{u}) du = [2u^2 + 24u + 36\ln(u)]$ from $u = 2$ to $u = 4$ .
Evaluating Antiderivative: Evaluate the antiderivative at the upper and lower limits: $[2(4)^2 + 24(4) + 36\ln(4)] - [2(2)^2 + 24(2) + 36\ln(2)]$ .
Simplifying Expression: Simplify the expression: $[2(16) + 96 + 36\ln(4)] - [2(4) + 48 + 36\ln(2)] = [32 + 96 + 36\ln(4)] - [8 + 48 + 36\ln(2)]$ .
Combining Like Terms: Combine like terms: $32 + 96 + 36\ln(4)$ - $8 + 48 + 36\ln(2)$ = $32 + 96 - 8 - 48$ + $36$ \ln( $4$ ) - $36$ \ln( $2$ ).
Further Simplification: Further simplify: $120 + 36\ln(4) - 36\ln(2) = 120 + 36(2\ln(2)) - 36\ln(2) = 120 + 72\ln(2) - 36\ln(2)$ .
Combining Logarithmic Terms: Combine the logarithmic terms: $120 + 36\ln(2) = a + b \ln(2)$ , where $a = 120$ and $b = 36$ .

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