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

The integral of x/(x^2+2x+2) is 1/2 ln|x^2+2x+2|-arctan(x+1)+C

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

\int \frac{x}{x ^{2} + 2 x + 2} d x

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

Solution steps

\int \frac{x}{x ^{2} + 2 x + 2} d x

Complete the square

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

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

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 1} d u

= \int \frac{u - 1}{u ^{2} + 1} d u

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{u ^{2} + 1} d u - \int \frac{1}{u ^{2} + 1} d u

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

Y (x) = x^{3} - x^{2} - 6 x, Y (x) = 0

f (x) = 7 x

\int e^{- 0.8 x} d x

\frac{\partial}{\partial x} (x^{2} \cdot ln (y))

\frac{d}{d x} (sin (2 x - \frac{π}{3}))

What is the integral of x/(x^2+2x+2) ?
The integral of x/(x^2+2x+2) is 1/2 ln|x^2+2x+2|-arctan(x+1)+C