What is the integral of 1/(sqrt(x^2+4x+20)) ?

The integral of 1/(sqrt(x^2+4x+20)) is ln(1/4 |x+2+sqrt(x^2+4x+20)|)+C

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integral of 1/(sqrt(x^2+4x+20))

\int \frac{1}{x ^{2} + 4 x + 20} d x

ln (\frac{1}{4} x + 2 + x^{2} + 4 x + 20) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 20} d x

Complete the square

x^{2} + 4 x + 20 : (x + 2)^{2} + 16

= \int \frac{1}{( x + 2 ) ^{2} + 16} d x

Apply u-substitution

= \int \frac{1}{u ^{2} + 16} d u

Apply Trigonometric Substitution

= \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}{4} (x + 2))) + sec (arctan (\frac{1}{4} (x + 2)))

Simplify

ln tan (arctan (\frac{1}{4} (x + 2))) + sec (arctan (\frac{1}{4} (x + 2))) : ln (\frac{1}{4} x + 2 + x^{2} + 4 x + 20)

= ln (\frac{1}{4} x + 2 + x^{2} + 4 x + 20)

Add a constant to the solution

= ln (\frac{1}{4} x + 2 + x^{2} + 4 x + 20) + C

d er i v a t i v e f (t) = t sin (π t)

\int x^{2} tan (x^{3}) d x

\frac{d ^{2}}{d x ^{2}} (5 cos (2 x + 10))

q \to 18 lim (0.9 q^{2} - 6 q + 80)

d er i v a t i v e sec (θ) (θ - tan (θ))

What is the integral of 1/(sqrt(x^2+4x+20)) ?
The integral of 1/(sqrt(x^2+4x+20)) is ln(1/4 |x+2+sqrt(x^2+4x+20)|)+C