What is the integral of 1/(sqrt(x^2+4x+13)) ?

The integral of 1/(sqrt(x^2+4x+13)) is ln(1/3 |x+2+sqrt(x^2+4x+13)|)+C

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integral of 1/(sqrt(x^2+4x+13))

\int \frac{1}{x ^{2} + 4 x + 13} d x

ln (\frac{1}{3} x + 2 + x^{2} + 4 x + 13) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= \int \frac{1}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} 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}{3} (x + 2))) + sec (arctan (\frac{1}{3} (x + 2)))

Simplify

ln tan (arctan (\frac{1}{3} (x + 2))) + sec (arctan (\frac{1}{3} (x + 2))) : ln (\frac{1}{3} x + 2 + x^{2} + 4 x + 13)

= ln (\frac{1}{3} x + 2 + x^{2} + 4 x + 13)

Add a constant to the solution

= ln (\frac{1}{3} x + 2 + x^{2} + 4 x + 13) + C

y^{^{''}} + 9 y = 5 sec (3 t)

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

y = cos (a^{7} + x^{7})

\frac{d}{d t} (e^{- t} cos (t))

\frac{d}{d x} (3 sin (π x))

What is the integral of 1/(sqrt(x^2+4x+13)) ?
The integral of 1/(sqrt(x^2+4x+13)) is ln(1/3 |x+2+sqrt(x^2+4x+13)|)+C