What is the integral of 1/(sqrt(x^2+4x+10)) ?

The integral of 1/(sqrt(x^2+4x+10)) is ln| 1/(sqrt(6))(x+2)+sqrt(1/6 (x^2+4x+10))|+C

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integral of 1/(sqrt(x^2+4x+10))

\int \frac{1}{x ^{2} + 4 x + 10} d x

ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 10} d x

Complete the square

x^{2} + 4 x + 10 : (x + 2)^{2} + 6

= \int \frac{1}{( x + 2 ) ^{2} + 6} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 6} d u

= \int \frac{1}{u ^{2} + 6} 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}{6} (x + 2))) + sec (arctan (\frac{1}{6} (x + 2)))

Simplify

ln tan (arctan (\frac{1}{6} (x + 2))) + sec (arctan (\frac{1}{6} (x + 2))) : ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10)

= ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10)

Add a constant to the solution

= ln \frac{1}{6} (x + 2) + \frac{1}{6} (x^{2} + 4 x + 10) + C

y cos^{2} (x) \frac{d y}{d x} = 4 + tan^{2} (x)

\int x + 7 d x

2 x^{3} - 3 x^{2} - 72 x + 108

ln (\frac{( 4 - x )}{4 + x})

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

What is the integral of 1/(sqrt(x^2+4x+10)) ?
The integral of 1/(sqrt(x^2+4x+10)) is ln| 1/(sqrt(6))(x+2)+sqrt(1/6 (x^2+4x+10))|+C