The integral ∫(1+x-1/x)e^{x+1/x}dx is equal to

Correct Answer is : xex+1x+cxe^{x+1/x}+cxex+x1+c

Correct Answer is : xex+1x+cxe^{x+1/x}+cxex+x1+c

Test Index

AP DSC School Assistant Math Model Test 1

Section: Subject Specific

Question : 52 of 160

Marks: +1, -0

The integral

∫(1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx

is equal to

Solution:

Concept:

This integral can be solved by rewriting the integrand to apply integration by parts, using the fact that the derivative of

x+\frac{1}{x}

1-\frac{1}{x^2}

Explanation:

Let

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

Split the integrand:

I = \int e^{x+\frac{1}{x}} \, dx + \int x\left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} \, dx

Notice that

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

. So the second integral is

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

Apply integration by parts:

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

Thus

I = \int e^{x+\frac{1}{x}} \, dx + x e^{x+\frac{1}{x}} - \int e^{x+\frac{1}{x}} \, dx = x e^{x+\frac{1}{x}} + c

Answer:

Option B:

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

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