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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

= \int \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

Apply u-substitution

: \int - e^{v} d v

= \int - e^{v} d v

Take the constant out:

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

= - \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= - e^{v}

Substitute back

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

Add a constant to the solution

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

\int \frac{ln ^{3} ( 1 - x )}{1 - x} d x

\int (x^{5} + 4 x^{2}) (x^{3} + 1)^{12} d x

\int \frac{35 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

\int \frac{9 x ^{2}}{( 16 + x ^{2} ) ^{2}} d x

\int x 1 5^{x} d x

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