Explanation for the correct option:Finding the given integral:[ ∫e^ logx dx = ∫x^ 1dx (∵e^nlogm=m^n); ...

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Integration by Substitution

∫e-logx dx =

Question

$\int e^{- \log x} d x =$

$e^{- \log x} + c$

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$x e^{- \log x} + c$

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$e^{\log x} + c$

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$\log | x | + c$

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Solution

The correct option is D
$\log | x | + c$

Explanation for the correct option:
Finding the given integral:
$\begin{array}{rcl} \int e^{- \log x} d x & = & \int x^{- 1} d x (∵ e^{n \log m} = m^{n}) \\ = & \int \frac{1}{x} d x (∵ \int \frac{1}{x} d x = \log |x|) \\ = & \log |x| + c \end{array}$
Hence, the correct answer is option (D).

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I_{1} = \int_{0}^{1} e^{- x} {cos}^{2} x d x, I_{2} = \int_{0}^{1} e^{- x^{2}} {cos}^{2} x d x

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The general solution of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q_{1}$ is

(a) $y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C$

(b) $y e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C$

(d) $x e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C$

\int e^{- 3 x} d x

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