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

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

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

\int \frac{2 x + 1}{x ^{2} + 2 x + 2} d x

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

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 2} d x

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

Substitute back

u = x + 1

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

Expand

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

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

Add a constant to the solution

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

\frac{d}{d x} (3 sin (x) - 3 x cos (x))

\frac{d}{d x} (cos^{x} (x))

\int 5 x + \frac{1}{2 x} d x

y = sin (x^{4})

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

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