What is the integral of (x+1)/(x^2+4x+13) ?

The integral of (x+1)/(x^2+4x+13) is 1/2 ln|x^2+4x+13|-1/3 arctan(1/3 (x+2))+C

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integral of (x+1)/(x^2+4x+13)

\int \frac{x + 1}{x ^{2} + 4 x + 13} d x

\frac{1}{2} ln x^{2} + 4 x + 13 - \frac{1}{3} arctan (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{x + 1}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int \frac{x + 1}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 9} d u

= \int \frac{u - 1}{u ^{2} + 9} d u

Expand

\frac{u - 1}{u ^{2} + 9} : \frac{u}{u ^{2} + 9} - \frac{1}{u ^{2} + 9}

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{u}{u ^{2} + 9} d u - \int \frac{1}{u ^{2} + 9} d u

\int \frac{u}{u ^{2} + 9} d u = \frac{1}{2} ln u^{2} + 9

\int \frac{1}{u ^{2} + 9} d u = \frac{1}{3} arctan (\frac{u}{3})

= \frac{1}{2} ln u^{2} + 9 - \frac{1}{3} arctan (\frac{u}{3})

Substitute back

u = x + 2

= \frac{1}{2} ln (x + 2)^{2} + 9 - \frac{1}{3} arctan (\frac{x + 2}{3})

Simplify

\frac{1}{2} ln (x + 2)^{2} + 9 - \frac{1}{3} arctan (\frac{x + 2}{3}) : \frac{1}{2} ln x^{2} + 4 x + 13 - \frac{1}{3} arctan (\frac{1}{3} (x + 2))

= \frac{1}{2} ln x^{2} + 4 x + 13 - \frac{1}{3} arctan (\frac{1}{3} (x + 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 13 - \frac{1}{3} arctan (\frac{1}{3} (x + 2)) + C

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

\int 2 t ln (t) d t

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

x^{3}

\int \frac{1}{- x ^{2} + 4 x - 5} d x

What is the integral of (x+1)/(x^2+4x+13) ?
The integral of (x+1)/(x^2+4x+13) is 1/2 ln|x^2+4x+13|-1/3 arctan(1/3 (x+2))+C