What is the integral of (2x-3)/(x^2+4x+5) ?

The integral of (2x-3)/(x^2+4x+5) is ln|x^2+4x+5|-7arctan(x+2)+C

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integral of (2x-3)/(x^2+4x+5)

\int \frac{2 x - 3}{x ^{2} + 4 x + 5} d x

ln x^{2} + 4 x + 5 - 7 arctan (x + 2) + C

Solution steps

\int \frac{2 x - 3}{x ^{2} + 4 x + 5} d x

Complete the square

x^{2} + 4 x + 5 : (x + 2)^{2} + 1

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

Apply u-substitution

: \int \frac{2 u - 7}{u ^{2} + 1} d u

= \int \frac{2 u - 7}{u ^{2} + 1} d u

Expand

\frac{2 u - 7}{u ^{2} + 1} : \frac{2 u}{u ^{2} + 1} - \frac{7}{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{7}{u ^{2} + 1} d u

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

\int \frac{7}{u ^{2} + 1} d u = 7 arctan (u)

= ln u^{2} + 1 - 7 arctan (u)

Substitute back

u = x + 2

= ln (x + 2)^{2} + 1 - 7 arctan (x + 2)

Expand

(x + 2)^{2} + 1 : x^{2} + 4 x + 5

= ln x^{2} + 4 x + 5 - 7 arctan (x + 2)

Add a constant to the solution

= ln x^{2} + 4 x + 5 - 7 arctan (x + 2) + C

n = 2 \sum \infty \frac{- 1}{n}

\frac{d}{d x} (\frac{x ^{3} - 2 x ^{2} + 2 x - 4}{x ^{2} + 2})

f (x) = \frac{c ^{9} - x ^{9}}{c ^{9} + x ^{9}}

\frac{d}{d x} (1 + 36 x)

\int \frac{1}{x ^{2} - 12} d x

What is the integral of (2x-3)/(x^2+4x+5) ?
The integral of (2x-3)/(x^2+4x+5) is ln|x^2+4x+5|-7arctan(x+2)+C