≤ft(1+x-1/x) e^{x+1/x} d x is equal to

Correct Answer is : xex+1x+c′x e^{x+1/x}+c^{}xex+x1+c′ where ccc is a constant of integration.

(x-1) e^{x+1/x}+c, where c is a constant of integration.

x e^{x+1/x}+c^{} where c is a constant of integration.

(x+1) e^{x+1/x}+c, where c is a constant of integration.

-x e^{x+1/x}+c^{} where c is a constant of integration.

Correct Answer is : xex+1x+c′x e^{x+1/x}+c^{}xex+x1+c′ where ccc is a constant of integration.

Test Index

MHTCET 15 May 2024 Shift 2 Paper

Question : 119 of 150

Marks: +1, -0

\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x

is equal to

Solution:

Concept:

Recognise the integrand as the derivative of

x e^{x+\frac{1}{x}}

Explanation:

We are given:

\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} dx

First, note that

\frac{d}{dx}\left(x+\frac{1}{x}\right) = 1 - \frac{1}{x^2}

Now, observe that

1+x-\frac{1}{x} = x\left(1-\frac{1}{x^2}\right) + 1

So, the integrand becomes:

\left[x\left(1-\frac{1}{x^2}\right) + 1\right] e^{x+\frac{1}{x}} = x e^{x+\frac{1}{x}}\left(1-\frac{1}{x^2}\right) + e^{x+\frac{1}{x}}

This is exactly of the form

x f'(x) + f(x)

, where

f(x) = e^{x+\frac{1}{x}}

Using the standard result:

\int \left[x f'(x) + f(x)\right] dx = x f(x) + c

, we get:

\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} dx = x e^{x+\frac{1}{x}} + c

Answer:

x e^{x+\frac{1}{x}} + c

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