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

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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

= - \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

x \to 1 + lim (x^{2} - 4 x + 3)

\frac{\partial}{\partial x} (4 e^{\frac{x ^{2}}{y}} - cos (x y^{3}))

x \to 0 lim (4 x + x)

x, y = x^{2}, x = y^{2}

\int 2 x^{7} + 4 x^{6} + x^{5} + 1 d x

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