Let tan^0 x = 1. If int left( sum_{k=0}^7 tan | vedclass

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(B) Given, $	an^0 x = 1$. We need to evaluate $\int \left( \sum_{k=0}^7 	an^k x ight) dx = \sum_{k=0}^7 \int 	an^k x dx$.Let $I_n = \int 	an^n x

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(B) Given, $	an^0 x = 1$. We need to evaluate $\int \left( \sum_{k=0}^7 	an^k x ight) dx = \sum_{k=0}^7 \int 	an^k x dx$.Let $I_n = \int 	an^n x dx$. We know the reduction formula: $I_n + I_{n-2} = \frac{	an^{n-1} x}{n-1}$.Expanding the sum: $\sum_{k=0}^7 I_k = I_0 + I_1 + I_2 + I_3 + I_4 + I_5 + I_6 + I_7$.Using the reduction formula:$I_0 + I_2 = 	an x$$I_1 + I_3 = \frac{	an^2 x}{2}$$I_2 + I_4 = \frac{	an^3 x}{3}$$I_3 + I_5 = \frac{	an^4 x}{4}$$I_4 + I_6 = \frac{	an^5 x}{5}$$I_5 + I_7 = \frac{	an^6 x}{6}$Note that $\int 	an^0 x dx = x$ and $\int 	an^1 x dx = \ln|\sec x|$. However, the problem states the integral is equal to $\sum_{k=1}^7 A_k 	an^k x + C$. This implies the terms involving $x$ and $\ln|\sec x|$ must cancel out or be zero. Assuming the question implies the polynomial part of the integral:Summing the reduction relations: $(I_0 + I_2) + (I_1 + I_3) + (I_2 + I_4) + (I_3 + I_5) + (I_4 + I_6) + (I_5 + I_7) = 	an x + \frac{	an^2 x}{2} + \frac{	an^3 x}{3} + \frac{	an^4 x}{4} + \frac{	an^5 x}{5} + \frac{	an^6 x}{6}$.Comparing this with $\sum_{k=1}^7 A_k 	an^k x$, we get $A_1 = 1, A_2 = 1/2, A_3 = 1/3, A_4 = 1/4, A_5 = 1/5, A_6 = 1/6, A_7 = 0$.Summing these coefficients: $\sum_{k=1}^7 A_k = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{60+30+20+15+12+10}{60} = \frac{147}{60} = \frac{49}{20}$.Re-evaluating the provided solution logic: The original solution provided in the prompt had a calculation error. Based on the standard reduction formula $\int 	an^n x dx = \frac{	an^{n-1} x}{n-1} - \int 	an^{n-2} x dx$, the sum of the coefficients is $1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 2.45 = 49/20$.

Let $\tan^0 x = 1$. If $\int \left( \sum_{k=0}^7 \tan^k x \right) dx = \sum_{k=1}^7 A_k \tan^k x + C$, then $\sum_{k=1}^7 A_k$ is equal to

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