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

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

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

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

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

Solution steps

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Apply Integration By Parts

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

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

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

Simplify

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

Add a constant to the solution

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

t an g e n t f (x) = x^{2} + 5 x - 3, at x = - 3

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

\frac{d}{d x} (\frac{3}{x ^{4} + 2})

t an g e n t f (x) = x^{2} - \frac{1}{x}, (2, 3.5)

n = 1 \sum \infty \frac{( ( - 7 ) ^{n - 1} )}{8 ^{n}}

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