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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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{2}{u ^{2} + 4} d u

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

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

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

Substitute back

u = x + 2

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

Simplify

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

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

Add a constant to the solution

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

\int_{- 2}^{2} ∣ x ∣ d x

\frac{6}{s ( 2 s ^{2} + 7 s + 3 )}

\frac{( s ^{2} - 1 )}{( s ^{2} + 1 ) ^{2}}

3 x^{2} e^{x} + e^{x} (x^{3} + 3)

arctan (3 x)

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