If ∫x(a+x)5dx=1k(a+x)4(f(x))+cthenf(−a)ak=

=∫x(a+x)5dx a+x=t dx=dt =∫t-at5dt =∫1t4dt-at5dt =-13t3+a4t4+c=112(a+x)4-4a+x+3a+f(−a)ak=14

If ∫x(a+x)5dx=1k(a+x)4(f(x))+cthenf(−a)ak=

=∫x(a+x)5dx a+x=t dx=dt =∫t-at5dt =∫1t4dt-at5dt =-13t3+a4t4+c=112(a+x)4-4a+x+3a+f(−a)ak=14

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If $\int \frac{x}{(a + x)^{5}} d x = \frac{1}{k (a + x)^{4}} (f (x)) + c then \frac{f (- a)}{a k} =$

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$\frac{1}{3}$

$\frac{1}{2}$

$\frac{5}{6}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

$= \int \frac{x}{(a + x)^{5}} d x a + x = t d x = d t = \int \frac{t - a}{t^{5}} d t = \int \frac{1}{t^{4}} d t - \frac{a}{t^{5}} d t = \frac{- 1}{3 t^{3}} + \frac{a}{4 t^{4}} + c$

$= \frac{1}{12 (a + x)^{4}} [- 4 (a + x) + 3 a] +$

$\frac{f (- a)}{a k} = \frac{1}{4}$

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If $\int \frac{x}{(a + x)^{5}} d x = \frac{1}{k (a + x)^{4}} (f (x)) + c then \frac{f (- a)}{a k} =$