If ln -1- x - x -1 e x - x -1 dx -A- x -1- e x - x -1- CB- x - e x - x -1- CC- x -1 - e x - x -1- CD- x 2-1 - e x - x -1- C

The correct option is B x. e^x+x^ 1+CI=∫ (1+x x^ 1)e^x+x^ 1 dx =∫ e^x+x^ 1 dx + ∫ x ( 1 1/x^2 )e^x+x^ 1 dx =∫ e^x+x^ 1 dx+x.e^x+x^ 1 ∫ e^x+x^ 1 dx =x.e^x ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

If ln =∫1+ x ...

Question

If $l n = \int (1 + x + x^{- 1}) e^{x + x^{- 1}} d x =$

x^{- 1} . e^{x + x^{- 1}} + C

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x . e^{x + x^{- 1}} + C

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(x + 1) . e^{x + x^{- 1}} + C

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(x^{2} + 1) . e^{x + x^{- 1}} + C

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Solution

The correct option is B $x . e^{x + x^{- 1}} + C$
$I = \int (1 + x - x^{- 1}) e^{x + x^{- 1}} d x = \int e^{x + x^{- 1}} d x + \int x (1 - \frac{1}{x^{2}}) e^{x + x^{- 1}} d x = \int e^{x + x^{- 1}} d x + x . e^{x + x^{- 1}} - \int e^{x + x^{- 1}} d x = x . e^{x + x^{- 1}} + c$

Suggest Corrections

If ln=∫(1+x+x−1)ex+x−1dx=

The correct option is B x.ex+x−1+CI=∫(1+x−x−1)ex+x−1 dx=∫ex+x−1 dx+∫x(1−1x2)ex+x−1 dx=∫ex+x−1 dx+x.ex+x−1−∫ex+x−1 dx=x.ex+x−1+c

If $l n = \int (1 + x + x^{- 1}) e^{x + x^{- 1}} d x =$

The correct option is B $x . e^{x + x^{- 1}} + C$
$I = \int (1 + x - x^{- 1}) e^{x + x^{- 1}} d x = \int e^{x + x^{- 1}} d x + \int x (1 - \frac{1}{x^{2}}) e^{x + x^{- 1}} d x = \int e^{x + x^{- 1}} d x + x . e^{x + x^{- 1}} - \int e^{x + x^{- 1}} d x = x . e^{x + x^{- 1}} + c$