What is the integral of (-11x-24)/(125(x^2+4x+5)) ?

The integral of (-11x-24)/(125(x^2+4x+5)) is 1/250 (-11ln|x^2+4x+5|-4arctan(x+2))+C

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integral of (-11x-24)/(125(x^2+4x+5))

\int \frac{- 11 x - 24}{125 ( x ^{2} + 4 x + 5 )} d x

\frac{1}{250} (- 11 ln x^{2} + 4 x + 5 - 4 arctan (x + 2)) + C

Solution steps

\int \frac{- 11 x - 24}{125 ( x ^{2} + 4 x + 5 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{125} \cdot \int \frac{- 11 x - 24}{x ^{2} + 4 x + 5} d x

Expand

\frac{- 11 x - 24}{x ^{2} + 4 x + 5} : - \frac{11 x}{x ^{2} + 4 x + 5} - \frac{24}{x ^{2} + 4 x + 5}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{125} (- \int \frac{11 x}{x ^{2} + 4 x + 5} d x - \int \frac{24}{x ^{2} + 4 x + 5} d x)

\int \frac{11 x}{x ^{2} + 4 x + 5} d x = 11 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2))

\int \frac{24}{x ^{2} + 4 x + 5} d x = 24 arctan (x + 2)

= \frac{1}{125} (- 11 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) - 24 arctan (x + 2))

Simplify

\frac{1}{125} (- 11 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) - 24 arctan (x + 2)) : \frac{1}{250} (- 11 ln x^{2} + 4 x + 5 - 4 arctan (x + 2))

= \frac{1}{250} (- 11 ln x^{2} + 4 x + 5 - 4 arctan (x + 2))

Add a constant to the solution

= \frac{1}{250} (- 11 ln x^{2} + 4 x + 5 - 4 arctan (x + 2)) + C

n = 1 \sum \infty \frac{2}{5 ^{n + 2}}

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

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

\int \frac{ln ( t )}{t ^{3}} d t

What is the integral of (-11x-24)/(125(x^2+4x+5)) ?
The integral of (-11x-24)/(125(x^2+4x+5)) is 1/250 (-11ln|x^2+4x+5|-4arctan(x+2))+C