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

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

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

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

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

Solution steps

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

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} (x + 1))) + sec (arctan (\frac{1}{2} (x + 1)))

Simplify

ln tan (arctan (\frac{1}{2} (x + 1))) + sec (arctan (\frac{1}{2} (x + 1))) : ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

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

Add a constant to the solution

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

y = 6 + 4 x^{2} - 2 x^{3}

\int_{1}^{\infty} 2 e^{- x} d x

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

n = 0 \sum \infty \frac{n}{e ^{n^{2}}}

\frac{\partial}{\partial y} (yz e^{x})

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