Question

If integral of (x^-3)5^(1/x^2)dx=k5^(1/x^2), then k is equal to?

Answer 1

SOLUTION

GIVEN

$\displaystyle \int\limits_{}^{} {x}^{ - 3} \: {5}^{ \frac{1}{ {x}^{2} } } \, dx = k \times {5}^{ \frac{1}{ {x}^{2} } } + c$

TO DETERMINE

The value of k

EVALUATION

Here the given Integral

$\displaystyle \int\limits_{}^{} {x}^{ - 3} \: {5}^{ \frac{1}{ {x}^{2} } } \, dx$

Let

$\displaystyle \: y = \frac{1}{ {x}^{2} }$

$\implies\displaystyle \: y = {x}^{ - 2}$

Differentiating both sides with respect to x we get

$\displaystyle \: \frac{dy}{dx} = - 2 {x}^{ - 3}$

$\implies\displaystyle \: - \frac{1}{2} dy = {x}^{ - 3} dx$

$\therefore \: \: \displaystyle \int\limits_{}^{} {x}^{ - 3} \: {5}^{ \frac{1}{ {x}^{2} } } \, dx$

$\: \: \displaystyle = - \frac{1}{2} \int\limits_{}^{} \: {5}^{ y } \, dy$

$\: \: \displaystyle = - \frac{1}{2} \times \frac{ {5}^{y} }{ \ln 5} + c$

$\: \: \displaystyle = - \frac{1}{2 \ln 5} \times {5}^{ \frac{1}{ {x}^{2} } } + c$

$\therefore \: \displaystyle - \frac{1}{2 \ln 5} \times {5}^{ \frac{1}{ {x}^{2} } } + c = k \times {5}^{ \frac{1}{ {x}^{2} } } + c$

Comparing both sides we get

$\: \displaystyle k = - \frac{1}{2 \ln 5}$

FINAL ANSWER

$\: \displaystyle k = - \frac{1}{2 \ln 5}$

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