If f(x)=x5+5x−1 then ∫541dx(f−1(x))5+5.f−1(x) equals

f(x)=x5+5x−1f(1)=5⇒f−1(5)=1f(2)=41⇒f−1(41)=2 Put f−1(x)=t ∴I=∫12f'(t)t5+5tdt=∫125t4+5t5+5tdt =ln(t5+5t)12=ln42−ln6=ln7

If f(x)=x5+5x−1 then ∫541dx(f−1(x))5+5.f−1(x) equals

f(x)=x5+5x−1f(1)=5⇒f−1(5)=1f(2)=41⇒f−1(41)=2 Put f−1(x)=t ∴I=∫12f'(t)t5+5tdt=∫125t4+5t5+5tdt =ln(t5+5t)12=ln42−ln6=ln7

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If $f (x) = x^{5} + 5 x - 1$ then $\int_{5}^{41} \frac{d x}{{(f^{- 1} (x))}^{5} + 5. f^{- 1} (x)}$ equals

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$\ln 7$

$0$

$\ln (\frac{41}{5})$

$\ln 2$

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = x^{5} + 5 x - 1 \\ f (1) = 5 \Rightarrow f^{- 1} (5) = 1 \\ f (2) = 41 \Rightarrow f^{- 1} (41) = 2 \end{array} P u t f^{- 1} (x) = t ∴ I = \int_{1}^{2} \frac{f' (t)}{t^{5} + 5 t} d t = \int_{1}^{2} \frac{5 t^{4} + 5}{t^{5} + 5 t} d t = \ln {(t^{5} + 5 t)}_{1}^{2} = \ln 42 - \ln 6 = \ln 7$