What is the integral of (x+1)/((x^2+4x+8)) ?

The integral of (x+1)/((x^2+4x+8)) is 1/2 ln|x^2+4x+8|-1/2 arctan(1/2 (x+2))+C

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integral of (x+1)/((x^2+4x+8))

integral

\frac{x + 1}{( x ^{2} + 4 x + 8 )}

\frac{1}{2} ln x^{2} + 4 x + 8 - \frac{1}{2} arctan (\frac{1}{2} (x + 2)) + C

Solution steps

\int \frac{x + 1}{x ^{2} + 4 x + 8} d x

Complete the square

x^{2} + 4 x + 8 : (x + 2)^{2} + 4

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

Apply u-substitution

: \int \frac{u - 1}{u ^{2} + 4} d u

= \int \frac{u - 1}{u ^{2} + 4} d u

Expand

\frac{u - 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{1}{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{u}{u ^{2} + 4} d u - \int \frac{1}{u ^{2} + 4} d u

\int \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln u^{2} + 4

\int \frac{1}{u ^{2} + 4} d u = \frac{1}{2} arctan (\frac{u}{2})

= \frac{1}{2} ln u^{2} + 4 - \frac{1}{2} arctan (\frac{u}{2})

Substitute back

u = x + 2

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

Simplify

\frac{1}{2} ln (x + 2)^{2} + 4 - \frac{1}{2} arctan (\frac{x + 2}{2}) : \frac{1}{2} ln x^{2} + 4 x + 8 - \frac{1}{2} arctan (\frac{1}{2} (x + 2))

= \frac{1}{2} ln x^{2} + 4 x + 8 - \frac{1}{2} arctan (\frac{1}{2} (x + 2))

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 8 - \frac{1}{2} arctan (\frac{1}{2} (x + 2)) + C

\frac{u}{( 1 + u )}

\frac{( x ^{3} - 1 ) ^{\frac{1}{2}}}{( x )}

\frac{x ^{4}}{4}

15 x^{2} (5 x^{3} - 6)^{5}

\frac{x ^{2}}{x ^{2} - 4}

What is the integral of (x+1)/((x^2+4x+8)) ?
The integral of (x+1)/((x^2+4x+8)) is 1/2 ln|x^2+4x+8|-1/2 arctan(1/2 (x+2))+C