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

The integral of (x+3)/(x^2+4x+5) is 1/2 ln|x^2+4x+5|+arctan(x+2)+C

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

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

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

Solution steps

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

Apply u-substitution

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

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

Complete the square

u^{2} - 2 u + 2 : (u - 1)^{2} + 1

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

Apply u-substitution

: \int \frac{v + 1}{v ^{2} + 1} d v

= \int \frac{v + 1}{v ^{2} + 1} d v

Expand

\frac{v + 1}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} + \frac{1}{v ^{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{v}{v ^{2} + 1} d v + \int \frac{1}{v ^{2} + 1} d v

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 5 + arctan (x + 2) + C

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What is the integral of (x+3)/(x^2+4x+5) ?
The integral of (x+3)/(x^2+4x+5) is 1/2 ln|x^2+4x+5|+arctan(x+2)+C