What is the integral of (5x+1)/(x^2+4x+5) ?

The integral of (5x+1)/(x^2+4x+5) is 5/2 ln|x^2+4x+5|-9arctan(x+2)+C

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integral of (5x+1)/(x^2+4x+5)

\int \frac{5 x + 1}{x ^{2} + 4 x + 5} d x

\frac{5}{2} ln x^{2} + 4 x + 5 - 9 arctan (x + 2) + C

Solution steps

\int \frac{5 x + 1}{x ^{2} + 4 x + 5} d x

Expand

\frac{5 x + 1}{x ^{2} + 4 x + 5} : \frac{5 x}{x ^{2} + 4 x + 5} + \frac{1}{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

= \int \frac{5 x}{x ^{2} + 4 x + 5} d x + \int \frac{1}{x ^{2} + 4 x + 5} d x

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

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

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

Simplify

5 (\frac{1}{2} ln x^{2} + 4 x + 5 - 2 arctan (x + 2)) + arctan (x + 2) : \frac{5}{2} ln x^{2} + 4 x + 5 - 9 arctan (x + 2)

= \frac{5}{2} ln x^{2} + 4 x + 5 - 9 arctan (x + 2)

Add a constant to the solution

= \frac{5}{2} ln x^{2} + 4 x + 5 - 9 arctan (x + 2) + C

\int_{0}^{5} \frac{2}{x ^{2}} d x

\frac{d y}{d t} = - \frac{y}{t} - 7

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

\frac{\partial}{\partial x} (z \cdot lo g_{e} (x^{2} + y^{2}))

(6 x^{4} - 9 x^{2} + 8)^{4}

What is the integral of (5x+1)/(x^2+4x+5) ?
The integral of (5x+1)/(x^2+4x+5) is 5/2 ln|x^2+4x+5|-9arctan(x+2)+C