The integral ∫ 1 + x - 1 x e x + 1 x d x is equal to

We have, I = ∫ 1 + x - 1 x e x + 1 x d x I = ∫ e x + 1 x d x + ∫ x - 1 x e x + 1 x d x I = ∫ e x + 1 x d x + ∫ x 1 - 1 x 2 e x + 1 x d x I = ∫ e x + 1 x d x + x e x + 1 x - ∫ e x + 1 x d x I = x e x + 1 x + C

Mathematics - Indefinite Integration Question with Solution | TestHub

MathematicsIndefinite IntegrationSubstitutionMedium2 minPYQ_2019

MathematicsMediumsingle choice

The integral $\int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} d x$ is equal to

Options:

Answer:

Solution:

We have,

$I = \int (1 + x - \frac{1}{x}) e^{x + \frac{1}{x}} d x$

$I = \int e^{x + \frac{1}{x}} d x + \int (x - \frac{1}{x}) e^{x + \frac{1}{x}} d x$

$I = \int e^{x + \frac{1}{x}} d x + \int x (1 - \frac{1}{x^{2}}) e^{x + \frac{1}{x}} d x$

$I = \int e^{x + \frac{1}{x}} d x + x e^{x + \frac{1}{x}} - \int e^{x + \frac{1}{x}} d x$

$I = x e^{x + \frac{1}{x}} + C$

Stream:BITSATSubject:MathematicsTopic:Indefinite IntegrationSubtopic:Substitution

⏱ 2mℹ️ Source: PYQ_2019

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