What is the integral of (2x+18)/(x^2+10x+24) ?

The integral of (2x+18)/(x^2+10x+24) is ln|x^2+10x+24|-8(1/2 ln|x+6|-1/2 ln|x+4|)+C

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integral of (2x+18)/(x^2+10x+24)

\int \frac{2 x + 18}{x ^{2} + 10 x + 24} d x

ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣) + C

Solution steps

\int \frac{2 x + 18}{x ^{2} + 10 x + 24} d x

Complete the square

x^{2} + 10 x + 24 : (x + 5)^{2} - 1

= \int \frac{2 x + 18}{( x + 5 ) ^{2} - 1} d x

Apply u-substitution

: \int \frac{2 u + 8}{u ^{2} - 1} d u

= \int \frac{2 u + 8}{u ^{2} - 1} d u

Expand

\frac{2 u + 8}{u ^{2} - 1} : \frac{2 u}{u ^{2} - 1} + \frac{8}{u ^{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{2 u}{u ^{2} - 1} d u + \int \frac{8}{u ^{2} - 1} d u

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

\int \frac{8}{u ^{2} - 1} d u = - 8 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

= ln u^{2} - 1 - 8 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

Substitute back

u = x + 5

= ln (x + 5)^{2} - 1 - 8 (\frac{1}{2} ln ∣ x + 5 + 1 ∣ - \frac{1}{2} ln ∣ x + 5 - 1 ∣)

Simplify

ln (x + 5)^{2} - 1 - 8 (\frac{1}{2} ln ∣ x + 5 + 1 ∣ - \frac{1}{2} ln ∣ x + 5 - 1 ∣) : ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣)

= ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣)

Add a constant to the solution

= ln x^{2} + 10 x + 24 - 8 (\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 4 ∣) + C

y = ln (9 + x^{2})

\frac{d}{d x} ((\frac{x ^{2} + 2 x}{sin ( x )})^{3})

\int x^{- 13} d x

\int \frac{1}{x ^{2} x ^{2} - 49} d x

\int (\frac{1}{x ( x - 4 )}) d x

What is the integral of (2x+18)/(x^2+10x+24) ?
The integral of (2x+18)/(x^2+10x+24) is ln|x^2+10x+24|-8(1/2 ln|x+6|-1/2 ln|x+4|)+C