What is the integral of ((1+1}{x^2})e^{\frac{x-1)/x} ?

The integral of ((1+1}{x^2})e^{\frac{x-1)/x} is 2e^{(x-1)/x}+C

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integral of ((1+1}{x^2})e^{\frac{x-1)/x}

integral

(\frac{1 + 1}{x ^{2}}) e^{\frac{x - 1}{x}}

2 e^{\frac{x - 1}{x}} + C

Solution steps

\int \frac{1 + 1}{x ^{2}} e^{\frac{x - 1}{x}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

= 2 \cdot \int \frac{1}{x ^{2}} e^{\frac{x - 1}{x}} d x

Apply u-substitution

= 2 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot u

Substitute back

u = e^{\frac{x - 1}{x}}

= 2 \cdot 1 \cdot e^{\frac{x - 1}{x}}

Simplify

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

Add a constant to the solution

= 2 e^{\frac{x - 1}{x}} + C

What is the integral of ((1+1}{x^2})e^{\frac{x-1)/x} ?
The integral of ((1+1}{x^2})e^{\frac{x-1)/x} is 2e^{(x-1)/x}+C