If f(x)=∫7x+x7−17xlog⁡7+7x6dx then f(1)−f(0)

f(x)=∫7xlog7+7x6x7+7xdx Let 7x+x7=t⇒7xlog⁡7+7x6dx=dtf(x)=∫ 1tdt=log⁡t+cf(x)=log⁡x7+7x+cf(1)−f(0)=log(8)−0=3log2

If f(x)=∫7x+x7−17xlog⁡7+7x6dx then f(1)−f(0)

f(x)=∫7xlog7+7x6x7+7xdx Let 7x+x7=t⇒7xlog⁡7+7x6dx=dtf(x)=∫ 1tdt=log⁡t+cf(x)=log⁡x7+7x+cf(1)−f(0)=log(8)−0=3log2

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$If f (x) = \int {(7^{x} + x^{7})}^{- 1} (7^{x} \log 7 + 7 x^{6}) d x then f (1) - f (0)$

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3 log 2

2 log 3

answer is B.

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$f (x) = \int \frac{7^{x} \log 7 + 7 x^{6}}{x^{7} + 7^{x}} d x$

$\begin{array}{l} Let 7^{x} + x^{7} = t \\ \Rightarrow (7^{x} \log 7 + 7 x^{6}) d x = d t \\ f (x) = \int_{}^{} \frac{1}{t} d t = \log t + c \\ f (x) = \log (x^{7} + 7^{x}) + c \end{array}$

$f (1) - f (0) = \log (8) - 0 = 3 \log 2$

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If f(x)=∫7x+x7−17xlog⁡7+7x6dx then f(1)−f(0)

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$If f (x) = \int {(7^{x} + x^{7})}^{- 1} (7^{x} \log 7 + 7 x^{6}) d x then f (1) - f (0)$