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

The integral of (2x-3)/(x^2+2x+2) is ln|x^2+2x+2|-5arctan(x+1)+C

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

\int \frac{2 x - 3}{x ^{2} + 2 x + 2} d x

ln x^{2} + 2 x + 2 - 5 arctan (x + 1) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} + 2 x + 2} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 5}{u ^{2} + 1} d u

= \int \frac{2 u - 5}{u ^{2} + 1} d u

Expand

\frac{2 u - 5}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{5}{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{5}{u ^{2} + 1} d u

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

\int \frac{5}{u ^{2} + 1} d u = 5 arctan (u)

= ln u^{2} + 1 - 5 arctan (u)

Substitute back

u = x + 1

= ln (x + 1)^{2} + 1 - 5 arctan (x + 1)

Expand

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

= ln x^{2} + 2 x + 2 - 5 arctan (x + 1)

Add a constant to the solution

= ln x^{2} + 2 x + 2 - 5 arctan (x + 1) + C

x \to - 6 lim (\frac{x ^{2} - 7}{6 - x})

\int_{0}^{0.5} 12 x - \frac{1}{12} x d x

\frac{\partial}{\partial x} (800 - 0.001 x^{2} - x y)

x \cdot 3^{x}

\frac{3 s}{s ^{2} - 2 s + 2}

What is the integral of (2x-3)/(x^2+2x+2) ?
The integral of (2x-3)/(x^2+2x+2) is ln|x^2+2x+2|-5arctan(x+1)+C