What is the integral of x/(sqrt(x^2+6x+8)) ?

The integral of x/(sqrt(x^2+6x+8)) is sqrt(x^2+6x+8)-3ln|sqrt(x^2+6x+8)+x+3|+C

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integral of x/(sqrt(x^2+6x+8))

\int \frac{x}{x ^{2} + 6 x + 8} d x

x^{2} + 6 x + 8 - 3 ln x^{2} + 6 x + 8 + x + 3 + C

Solution steps

\int \frac{x}{x ^{2} + 6 x + 8} d x

Complete the square

x^{2} + 6 x + 8 : (x + 3)^{2} - 1

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

Apply u-substitution

: \int \frac{u - 3}{u ^{2} - 1} d u

= \int \frac{u - 3}{u ^{2} - 1} d u

Expand

\frac{u - 3}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} - \frac{3}{u ^{2} - 1}

= \int \frac{u}{u ^{2} - 1} - \frac{3}{u ^{2} - 1} 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} - 1} d u - \int \frac{3}{u ^{2} - 1} d u

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

\int \frac{3}{u ^{2} - 1} d u = 3 ln u^{2} - 1 + u

= u^{2} - 1 - 3 ln u^{2} - 1 + u

Substitute back

u = x + 3

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

Simplify

(x + 3)^{2} - 1 - 3 ln (x + 3)^{2} - 1 + x + 3 : x^{2} + 6 x + 8 - 3 ln x^{2} + 6 x + 8 + x + 3

= x^{2} + 6 x + 8 - 3 ln x^{2} + 6 x + 8 + x + 3

Add a constant to the solution

= x^{2} + 6 x + 8 - 3 ln x^{2} + 6 x + 8 + x + 3 + C

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

\int 6 cot^{4} (x) d x

\int \frac{1}{t ^{3}} cos (\frac{π}{t ^{2}}) d t

y^{^{''}} - 4 y = e^{- 2 x} - 2 x, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y = \frac{3}{2} sin (x)

What is the integral of x/(sqrt(x^2+6x+8)) ?
The integral of x/(sqrt(x^2+6x+8)) is sqrt(x^2+6x+8)-3ln|sqrt(x^2+6x+8)+x+3|+C