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Question: \(\int_{}^{}{x^{- 3}5^{1/x^{2}}dx = k.5^{1/x^{2}} + c,}\) then k is...

∫x−351/x2dx=k.51/x2+c,\int_{}^{}{x^{- 3}5^{1/x^{2}}dx = k.5^{1/x^{2}} + c,} then k is

A

−12log⁡5+C- \frac{1}{2\log 5} + C

B

−2log⁡5- 2\log 5

C

2log⁡5\frac{2}{\log 5}

D

−2log⁡5\frac{- 2}{\log 5}

Answer

−12log⁡5+C- \frac{1}{2\log 5} + C

Explanation

Solution

Put x−2=tx^{- 2} = t ⇒−2x−3dx=dt\Rightarrow - 2x^{- 3}dx = dt ⇒x−3dx=−dt2\Rightarrow x^{- 3}dx = - \frac{dt}{2}

∫x−351/x2dx=−12∫5tdt=−125tlog⁡e5+c\int_{}^{}{x^{- 3}5^{1/x^{2}}dx} = - \frac{1}{2}\int_{}^{}5^{t}dt = - \frac{1}{2}\frac{5^{t}}{\log_{e}5} + c =12log⁡e5.51/x2+C= \frac{1}{2\log_{e}5}.5^{1/x^{2}} + C. On comparing, k=−12log⁡e5k = - \frac{1}{2\log_{e}5}