What is the integral of ((x+1))/((x^2+2x+2)^3) ?

The integral of ((x+1))/((x^2+2x+2)^3) is -1/(4(x^2+2x+2)^2)+C

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integral of ((x+1))/((x^2+2x+2)^3)

\int \frac{( x + 1 )}{( x ^{2} + 2 x + 2 ) ^{3}} d x

- \frac{1}{4 ( x ^{2} + 2 x + 2 ) ^{2}} + C

Solution steps

\int \frac{x + 1}{( x ^{2} + 2 x + 2 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= \int \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{2} + 2 x + 2

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

Simplify

\frac{1}{2} (- \frac{1}{2 ( x ^{2} + 2 x + 2 ) ^{2}}) : - \frac{1}{4 ( x ^{2} + 2 x + 2 ) ^{2}}

= - \frac{1}{4 ( x ^{2} + 2 x + 2 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( x ^{2} + 2 x + 2 ) ^{2}} + C

\frac{d y}{d x} = 1 + \frac{y}{x}

y = lo g_{4} (x) + lo g_{4} (x^{2})

\frac{d}{d x} (x^{2} - 8)

(0, 2), (2, - 3)

δ (x)

What is the integral of ((x+1))/((x^2+2x+2)^3) ?
The integral of ((x+1))/((x^2+2x+2)^3) is -1/(4(x^2+2x+2)^2)+C