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

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

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integral of (2x+5)/(x^2+2x+5)

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

ln x^{2} + 2 x + 5 + \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + C

Solution steps

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

Complete the square

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

= \int \frac{2 x + 5}{( x + 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{2 u + 3}{u ^{2} + 4} d u

= \int \frac{2 u + 3}{u ^{2} + 4} d u

Expand

\frac{2 u + 3}{u ^{2} + 4} : \frac{2 u}{u ^{2} + 4} + \frac{3}{u ^{2} + 4}

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} + 4} d u + \int \frac{3}{u ^{2} + 4} d u

\int \frac{2 u}{u ^{2} + 4} d u = ln u^{2} + 4

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

= ln u^{2} + 4 + \frac{3}{2} arctan (\frac{u}{2})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 4 + \frac{3}{2} arctan (\frac{x + 1}{2})

Simplify

ln (x + 1)^{2} + 4 + \frac{3}{2} arctan (\frac{x + 1}{2}) : ln x^{2} + 2 x + 5 + \frac{3}{2} arctan (\frac{1}{2} (x + 1))

= ln x^{2} + 2 x + 5 + \frac{3}{2} arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 5 + \frac{3}{2} arctan (\frac{1}{2} (x + 1)) + C

\int_{1}^{4} (x^{2} - 4 x - 3) d x

\int cos (6 θ) d θ

x \to \frac{1}{5} lim (\frac{x}{( 10 x - 2 )})

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

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

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