What is the integral of 1/(sqrt(x^2+6x+13)) ?

The integral of 1/(sqrt(x^2+6x+13)) is ln(1/2 |x+3+sqrt(x^2+6x+13)|)+C

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integral of 1/(sqrt(x^2+6x+13))

\int \frac{1}{x ^{2} + 6 x + 13} d x

ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

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

Simplify

ln tan (arctan (\frac{1}{2} (x + 3))) + sec (arctan (\frac{1}{2} (x + 3))) : ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13)

= ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13)

Add a constant to the solution

= ln (\frac{1}{2} x + 3 + x^{2} + 6 x + 13) + C

\int_{1}^{4} 3 e^{x} d x

\frac{d}{d x} (tan (π))

x = 1, y = 6 x (2 - x^{2})^{2}

x \to - 4 - lim (\frac{- 4 x ^{3}}{4 + x})

x \to a + lim (x^{n})

What is the integral of 1/(sqrt(x^2+6x+13)) ?
The integral of 1/(sqrt(x^2+6x+13)) is ln(1/2 |x+3+sqrt(x^2+6x+13)|)+C