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

The integral of (x+4)/(sqrt(x^2+2x+5)) is sqrt(x^2+2x+5)+3ln(1/2 |x+1+sqrt(x^2+2x+5)|)+C

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

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

x^{2} + 2 x + 5 + 3 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

= x^{2} + 2 x + 5 + 3 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

Add a constant to the solution

= x^{2} + 2 x + 5 + 3 ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

\frac{d}{d x} (\frac{- x ^{2} - 7}{( x ^{2} - 7 ) ^{2}})

x \to 0 + lim (x^{21 x})

\frac{\partial}{\partial y} (\frac{3 y ^{2} - t ^{2}}{y ^{5}})

\frac{d}{d x} ((x^{2} + 1)^{8})

\frac{( 2 s + 1 )}{s ( 2 s + 10 )}

What is the integral of (x+4)/(sqrt(x^2+2x+5)) ?
The integral of (x+4)/(sqrt(x^2+2x+5)) is sqrt(x^2+2x+5)+3ln(1/2 |x+1+sqrt(x^2+2x+5)|)+C